Answer: the series diverges.
Example No. 3
Find the sum of the series $\sum\limits_(n=1)^(\infty)\frac(2)((2n+1)(2n+3))$.
Since the lower limit of summation is 1, the common term of the series is written under the sum sign: $u_n=\frac(2)((2n+1)(2n+3))$. Let's make the nth partial sum of the series, i.e. Let's sum the first $n$ terms of a given number series:
$$ S_n=u_1+u_2+u_3+u_4+\ldots+u_n=\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\frac(2)(7\cdot 9 )+\frac(2)(9\cdot 11)+\ldots+\frac(2)((2n+1)(2n+3)). $$
Why I write exactly $\frac(2)(3\cdot 5)$, and not $\frac(2)(15)$, will be clear from the further narration. However, writing down a partial amount did not bring us one iota closer to our goal. We need to find $\lim_(n\to\infty)S_n$, but if we just write:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(2)(3\cdot 5)+\frac(2)(5\cdot 7)+\ frac(2)(7\cdot 9)+\frac(2)(9\cdot 11)+\ldots+\frac(2)((2n+1)(2n+3))\right), $$
then this record, completely correct in form, will give us nothing in essence. To find the limit, the expression for the partial sum must first be simplified.
There is a standard transformation for this, which consists in decomposing the fraction $\frac(2)((2n+1)(2n+3))$, which represents the general term of the series, into elementary fractions. A separate topic is devoted to the issue of decomposing rational fractions into elementary ones (see, for example, example No. 3 on this page). Expanding the fraction $\frac(2)((2n+1)(2n+3))$ into elementary fractions, we will have:
$$ \frac(2)((2n+1)(2n+3))=\frac(A)(2n+1)+\frac(B)(2n+3)=\frac(A\cdot(2n +3)+B\cdot(2n+1))((2n+1)(2n+3)). $$
We equate the numerators of the fractions on the left and right sides of the resulting equality:
$$ 2=A\cdot(2n+3)+B\cdot(2n+1). $$
There are two ways to find the values of $A$ and $B$. You can open the brackets and rearrange the terms, or you can simply substitute some suitable values instead of $n$. Just for variety, in this example we will go the first way, and in the next one we will substitute private values $n$. Opening the brackets and rearranging the terms, we get:
$$ 2=2An+3A+2Bn+B;\\ 2=(2A+2B)n+3A+B. $$
On the left side of the equality, $n$ is preceded by a zero. If you like, for clarity, the left side of the equality can be represented as $0\cdot n+ 2$. Since on the left side of the equality $n$ is preceded by zero, and on the right side of the equality $n$ is preceded by $2A+2B$, we have the first equation: $2A+2B=0$. Let's immediately divide both sides of this equation by 2, after which we get $A+B=0$.
Since on the left side of the equality the free term is equal to 2, and on the right side of the equality the free term is equal to $3A+B$, then $3A+B=2$. So, we have a system:
$$ \left\(\begin(aligned) & A+B=0;\\ & 3A+B=2. \end(aligned)\right. $$
We will carry out the proof using the method of mathematical induction. At the first step, you need to check whether the equality being proved is true $S_n=\frac(1)(3)-\frac(1)(2n+3)$ for $n=1$. We know that $S_1=u_1=\frac(2)(15)$, but will the expression $\frac(1)(3)-\frac(1)(2n+3)$ give the value $\frac(2 )(15)$, if we substitute $n=1$ into it? Let's check:
$$ \frac(1)(3)-\frac(1)(2n+3)=\frac(1)(3)-\frac(1)(2\cdot 1+3)=\frac(1) (3)-\frac(1)(5)=\frac(5-3)(15)=\frac(2)(15). $$
So, for $n=1$ the equality $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is satisfied. This completes the first step of the method of mathematical induction.
Let us assume that for $n=k$ the equality is satisfied, i.e. $S_k=\frac(1)(3)-\frac(1)(2k+3)$. Let us prove that the same equality will be satisfied for $n=k+1$. To do this, consider $S_(k+1)$:
$$ S_(k+1)=S_k+u_(k+1). $$
Since $u_n=\frac(1)(2n+1)-\frac(1)(2n+3)$, then $u_(k+1)=\frac(1)(2(k+1)+ 1)-\frac(1)(2(k+1)+3)=\frac(1)(2k+3)-\frac(1)(2(k+1)+3)$. According to the assumption made above $S_k=\frac(1)(3)-\frac(1)(2k+3)$, therefore the formula $S_(k+1)=S_k+u_(k+1)$ will take the form:
$$ S_(k+1)=S_k+u_(k+1)=\frac(1)(3)-\frac(1)(2k+3)+\frac(1)(2k+3)-\ frac(1)(2(k+1)+3)=\frac(1)(3)-\frac(1)(2(k+1)+3). $$
Conclusion: the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is correct for $n=k+1$. Therefore, according to the method of mathematical induction, the formula $S_n=\frac(1)(3)-\frac(1)(2n+3)$ is true for any $n\in N$. Equality has been proven.
In a standard course of higher mathematics, they are usually content with “crossing out” canceling terms, without requiring any proof. So, we got the expression for the nth partial sum: $S_n=\frac(1)(3)-\frac(1)(2n+3)$. Let's find the value of $\lim_(n\to\infty)S_n$:
Conclusion: the given series converges and its sum is $S=\frac(1)(3)$.
The second way to simplify the formula for a partial sum.
Honestly, I prefer this method myself :) Let's write down the partial amount in an abbreviated version:
$$ S_n=\sum\limits_(k=1)^(n)u_k=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)). $$
We obtained earlier that $u_k=\frac(1)(2k+1)-\frac(1)(2k+3)$, therefore:
$$ S_n=\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3))=\sum\limits_(k=1)^(n)\left (\frac(1)(2k+1)-\frac(1)(2k+3)\right). $$
The sum $S_n$ contains a finite number of terms, so we can rearrange them as we please. I want to first add all terms of the form $\frac(1)(2k+1)$, and only then move on to terms of the form $\frac(1)(2k+3)$. This means that we will present the partial amount as follows:
$$ S_n =\frac(1)(3)-\frac(1)(5)+\frac(1)(5)-\frac(1)(7)+\frac(1)(7)-\ frac(1)(9)+\frac(1)(9)-\frac(1)(11)+\ldots+\frac(1)(2n+1)-\frac(1)(2n+3)= \\ =\frac(1)(3)+\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+1 )-\left(\frac(1)(5)+\frac(1)(7)+\frac(1)(9)+\ldots+\frac(1)(2n+3)\right). $$
Of course, the expanded notation is extremely inconvenient, so the above equality can be written more compactly:
$$ S_n=\sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\sum\limits_( k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3). $$
Now let's transform the expressions $\frac(1)(2k+1)$ and $\frac(1)(2k+3)$ into one form. I think it is convenient to reduce it to the form of a larger fraction (although it is possible to use a smaller one, this is a matter of taste). Since $\frac(1)(2k+1)>\frac(1)(2k+3)$ (the larger the denominator, the smaller the fraction), we will give the fraction $\frac(1)(2k+3) $ to the form $\frac(1)(2k+1)$.
I will present the expression in the denominator of the fraction $\frac(1)(2k+3)$ as follows:
$$ \frac(1)(2k+3)=\frac(1)(2k+2+1)=\frac(1)(2(k+1)+1). $$
And the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+3)$ can now be written as follows:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2(k+1) )+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+1). $$
If the equality $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=2)^(n+1)\frac(1)(2k+ 1) $ does not raise any questions, then let's move on. If you have any questions, please expand the note.
How did we get the converted amount? show\hide
We had a series $\sum\limits_(k=1)^(n)\frac(1)(2k+3)=\sum\limits_(k=1)^(n)\frac(1)(2( k+1)+1)$. Let's introduce a new variable instead of $k+1$ - for example, $t$. So $t=k+1$.
How did the old variable $k$ change? And it changed from 1 to $n$. Let's find out how the new variable $t$ will change. If $k=1$, then $t=1+1=2$. If $k=n$, then $t=n+1$. So, the expression $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)$ now becomes: $\sum\limits_(t=2)^(n +1)\frac(1)(2t+1)$.
$$ \sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(t=2)^(n+1)\frac(1 )(2t+1). $$
We have the sum $\sum\limits_(t=2)^(n+1)\frac(1)(2t+1)$. Question: does it matter which letter is used in this amount? :) Simply writing the letter $k$ instead of $t$, we get the following:
$$ \sum\limits_(t=2)^(n+1)\frac(1)(2t+1)=\sum\limits_(k=2)^(n+1)\frac(1)(2k +1). $$
This is how we get the equality $\sum\limits_(k=1)^(n)\frac(1)(2(k+1)+1)=\sum\limits_(k=2)^(n+1) \frac(1)(2k+1)$.
Thus, the partial sum can be represented as follows:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k+1 ). $$
Note that the sums $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ and $\sum\limits_(k=2)^(n+1)\frac(1 )(2k+1)$ differ only in the summation limits. Let's make these limits the same. “Taking away” the first element from the sum $\sum\limits_(k=1)^(n)\frac(1)(2k+1)$ we will have:
$$ \sum\limits_(k=1)^(n)\frac(1)(2k+1)=\frac(1)(2\cdot 1+1)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1). $$
“Taking away” the last element from the sum $\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)$, we get:
$$\sum\limits_(k=2)^(n+1)\frac(1)(2k+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1 )+\frac(1)(2(n+1)+1)=\sum\limits_(k=2)^(n)\frac(1)(2k+1)+\frac(1)(2n+ 3).$$
Then the expression for the partial sum will take the form:
$$ S_n=\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n+1)\frac(1)(2k +1)=\frac(1)(3)+\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^ (n)\frac(1)(2k+1)+\frac(1)(2n+3)\right)=\\ =\frac(1)(3)+\sum\limits_(k=2)^ (n)\frac(1)(2k+1)-\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\frac(1)(2n+3)=\ frac(1)(3)-\frac(1)(2n+3). $$
If you skip all the explanations, then the process of finding a shortened formula for the nth partial sum will take the following form:
$$ S_n=\sum\limits_(k=1)^(n)u_k =\sum\limits_(k=1)^(n)\frac(2)((2k+1)(2k+3)) = \sum\limits_(k=1)^(n)\left(\frac(1)(2k+1)-\frac(1)(2k+3)\right)=\\ =\sum\limits_(k =1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3) =\frac(1)(3) +\sum\limits_(k=2)^(n)\frac(1)(2k+1)-\left(\sum\limits_(k=2)^(n)\frac(1)(2k+1 )+\frac(1)(2n+3)\right)=\frac(1)(3)-\frac(1)(2n+3). $$
Let me remind you that we reduced the fraction $\frac(1)(2k+3)$ to the form $\frac(1)(2k+1)$. Of course, you can do the opposite, i.e. represent the fraction $\frac(1)(2k+1)$ as $\frac(1)(2k+3)$. The final expression for the partial sum will not change. In this case, I will hide the process of finding the partial amount under a note.
How to find $S_n$ if converted to another fraction? show\hide
$$ S_n =\sum\limits_(k=1)^(n)\frac(1)(2k+1)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 ) =\sum\limits_(k=0)^(n-1)\frac(1)(2k+3)-\sum\limits_(k=1)^(n)\frac(1)(2k+3 )=\\ =\frac(1)(3)+\sum\limits_(k=1)^(n-1)\frac(1)(2k+3)-\left(\sum\limits_(k= 1)^(n-1)\frac(1)(2k+3)+\frac(1)(2n+3)\right) =\frac(1)(3)-\frac(1)(2n+ 3). $$
So, $S_n=\frac(1)(3)-\frac(1)(2n+3)$. Find the limit $\lim_(n\to\infty)S_n$:
$$ \lim_(n\to\infty)S_n=\lim_(n\to\infty)\left(\frac(1)(3)-\frac(1)(2n+3)\right)=\frac (1)(3)-0=\frac(1)(3). $$
The given series converges and its sum $S=\frac(1)(3)$.
Answer: $S=\frac(1)(3)$.
The continuation of the topic of finding the sum of a series will be discussed in the second and third parts.
Basic definitions
Definition. The sum of the terms of an infinite number sequence is called a number series.
In this case, we will call the numbers members of the series, and un - the common term of the series.
Definition. Sums, n = 1, 2, ... are called private (partial) sums of the series.
Thus, it is possible to consider sequences of partial sums of the series S1, S2, …, Sn, …
Definition. A series is called convergent if the sequence of its partial sums converges. The sum of a convergent series is the limit of the sequence of its partial sums.
Definition. If the sequence of partial sums of a series diverges, i.e. has no limit, or has an infinite limit, then the series is called divergent and no sum is assigned to it.
Row Properties
1) The convergence or divergence of the series will not be violated if you change, discard or add a finite number of terms of the series.
2) Consider two series and, where C is a constant number.
Theorem. If a series converges and its sum is equal to S, then the series also converges and its sum is equal to CS. (C 0)
3) Consider two rows and. The sum or difference of these series will be called a series where the elements are obtained as a result of the addition (subtraction) of the original elements with the same numbers.
Theorem. If the series and converge and their sums are equal to S and, respectively, then the series also converges and its sum is equal to S +.
The difference of two convergent series will also be a convergent series.
The sum of a convergent and a divergent series is a divergent series.
It is impossible to make a general statement about the sum of two divergent series.
When studying series, they mainly solve two problems: studying convergence and finding the sum of the series.
Cauchy criterion.
(necessary and sufficient conditions for the convergence of the series)
In order for a sequence to be convergent, it is necessary and sufficient that for any there exists a number N such that for n > N and any p > 0, where p is an integer, the inequality would hold:
Proof. (necessity)
Let then for any number there is a number N such that the inequality
is fulfilled when n>N. For n>N and any integer p>0 the inequality also holds. Taking into account both inequalities, we obtain:
The need has been proven. We will not consider the proof of sufficiency.
Let us formulate the Cauchy criterion for the series.
In order for a series to be convergent, it is necessary and sufficient that for any there exist a number N such that for n>N and any p>0 the inequality would hold
However, in practice, using the Cauchy criterion directly is not very convenient. Therefore, as a rule, simpler convergence tests are used:
1) If the series converges, then it is necessary that the common term un tends to zero. However, this condition is not sufficient. We can only say that if the common term does not tend to zero, then the series definitely diverges. For example, the so-called harmonic series is divergent, although its common term tends to zero.
A number series is a sequence that is considered together with another sequence (it is also called a sequence of partial sums). Similar concepts are used in mathematical and complex analysis.
The sum of a number series can be easily calculated in Excel using the SERIES.SUM function. Let's look at an example of how this function works, and then build a graph of the functions. Let's learn how to use the number series in practice when calculating capital growth. But first, a little theory.
Number series sum
The number series can be considered as a system of approximations to numbers. To designate it, use the formula:
Here is the initial sequence of numbers in the series and the summation rule:
- ∑ - mathematical sign of the sum;
- a i - general argument;
- i is a variable, a rule for changing each subsequent argument;
- ∞ is the infinity sign, the “limit” up to which the summation is carried out.
The notation means: natural numbers from 1 to “plus infinity” are summed. Since i = 1, the calculation of the sum starts from one. If there was another number here (for example, 2, 3), then we would start summing from it (from 2, 3).
In accordance with the variable i, the series can be written expanded:
A 1 + a 2 + a 3 + a 4 + a 5 + ... (up to “plus infinity”).
The definition of the sum of a number series is given through “partial sums”. In mathematics they are denoted Sn. Let's write our number series in the form of partial sums:
S 2 = a 1 + a 2
S 3 = a 1 + a 2 + a 3
S 4 = a 1 + a 2 + a 3 + a 4
The sum of a number series is the limit of partial sums S n . If the limit is finite, we speak of a “convergent” series. Infinite - about “divergent”.
First, let's find the sum of the number series:
Now let’s build a table of values of series members in Excel:
We take the general first argument from the formula: i=3.
We find all the following values of i using the formula: =B4+$B$1. Place the cursor in the lower right corner of cell B5 and multiply the formula.
Let's find the values. Make cell C4 active and enter the formula: =SUM(2*B4+1). Copy cell C4 to the specified range.
The value of the sum of arguments is obtained using the function: =SUM(C4:C11). Hotkey combination ALT+“+” (plus on the keyboard).
ROW.SUM function in Excel
To find the sum of a number series in Excel, use the mathematical function SERIES.SUM. The program uses the following formula:
Function arguments:
- x – variable value;
- n – degree for the first argument;
- m is the step by which the degree is increased for each subsequent term;
- a are the coefficients for the corresponding powers of x.
Important conditions for the function to work:
- all arguments are required (that is, all must be filled in);
- all arguments are NUMERIC values;
- the vector of coefficients has a fixed length (the limit of “infinity” will not work);
- number of “coefficients” = number of arguments.
Calculating the sum of a series in Excel
The same SERIES.SUM function works with power series (one of the variants of functional series). Unlike numeric ones, their arguments are functions.
Functional series are often used in the financial and economic sphere. You could say this is their application area.
For example, they deposited a certain amount of money (a) in the bank for a certain period (n). We have an annual payment of x percent. To calculate the accrued amount at the end of the first period, the formula is used:
S 1 = a (1 + x).
At the end of the second and subsequent periods, the form of expressions is as follows:
S 2 = a (1 + x) 2 ; S 3 = a (1 + x) 2, etc.
To find the total:
S n = a (1 + x) + a (1 + x) 2 + a (1 + x) 3 + … + a (1 + x) n
Partial sums in Excel can be found using the BS() function.
Initial parameters for the training task:
Using a standard mathematical function, we find the accumulated amount at the end of the term. To do this, in cell D2 we use the formula: =B2*DEGREE(1+B3;4)
Now in cell D3 we will solve the same problem using the built-in Excel function: =BS(B3;B1;;-B2)
The results are the same, as it should be.
How to fill the arguments of the BS() function:
- “Rate” is the interest rate at which the deposit is made. Since the percentage format is set in cell B3, we simply specified a link to this cell in the argument field. If a number were specified, then it would be written as a hundredth of it (20/100).
- “Nper” is the number of periods for interest payments. In our example – 4 years.
- "Plt" - periodic payments. In our case there are none. Therefore, we do not fill in the argument field.
- “Ps” - “present value”, the amount of the deposit. Since we are parting with this money for a while, we indicate the parameter with a “-” sign.
Thus, the BS function helped us find the sum of the functional series.
Excel has other built-in functions for finding different parameters. Typically these are functions for working with investment projects, securities and depreciation payments.
Plotting functions of the sum of a number series
Let's build a function graph reflecting capital growth. To do this, we need to construct a graph of a function that is the sum of the constructed series. As an example, let’s take the same data on the deposit:
The first line shows the accumulated amount after one year. In the second - in two. And so on.
Let's create another column in which we will reflect the profit:
As we thought - in the formula bar.
Based on the data obtained, we will construct a graph of functions.
Let's select 2 ranges: A5:A9 and C5:C9. Go to the “Insert” tab - the “Diagrams” tool. Select the first chart:
Let's make the problem even more “applied”. In the example we used compound interest. They are accrued on the amount accrued in the previous period.
Let's take simple interest for comparison. Simple interest formula in Excel: =$B$2*(1+A6*B6)
Let’s add the obtained values to the “Capital Growth” chart.
It is obvious what conclusions the investor will draw.
Mathematical formula for the partial sum of a functional series (with simple interest): S n = a (1 + x*n), where a is the initial deposit amount, x is interest, n is period.
In order to calculate the sum of a series, you just need to add the elements of the row a given number of times. For example:
In the example above, this was done very simply, since it had to be summed a finite number of times. But what if the upper limit of summation is infinity? For example, if we need to find the sum of the following series:
By analogy with the previous example, we can write this amount like this:
But what to do next?! At this stage it is necessary to introduce the concept partial sum of the series. So, partial sum of the series(denoted S n) is the sum of the first n terms of the series. Those. in our case:
Then the sum of the original series can be calculated as the limit of the partial sum:
Thus, for calculating the sum of a series, it is necessary to somehow find an expression for the partial sum of the series (S n ). In our particular case, the series is a decreasing geometric progression with a denominator of 1/3. As you know, the sum of the first n elements of a geometric progression is calculated by the formula:
here b 1 is the first element of the geometric progression (in our case it is 1) and q is the denominator of the progression (in our case 1/3). Therefore, the partial sum S n for our series is equal to:
Then the sum of our series (S) according to the definition given above is equal to:
The examples discussed above are quite simple. Usually, calculating the sum of a series is much more difficult and the greatest difficulty lies in finding the partial sum of the series. The online calculator presented below, created on the basis of the Wolfram Alpha system, allows you to calculate the sum of quite complex series. Moreover, if the calculator could not find the sum of a series, it is likely that the series is divergent (in which case the calculator displays a message like “sum diverges”), i.e. This calculator also indirectly helps to get an idea of the convergence of series.
To find the sum of your series, you need to specify the variable of the series, the lower and upper limits of the summation, as well as the expression for the nth term of the series (i.e., the actual expression for the series itself).
Basic definitions.
Definition. The sum of the terms of an infinite number sequence is called number series.
At the same time, the numbers 
we will call them members of the series, and u n– a common member of the series.
Definition.
Amounts 
,n
= 1, 2, …
are called private (partial) amounts row.
Thus, it is possible to consider sequences of partial sums of the series S 1 , S 2 , …, S n , …
Definition.
Row 
called convergent, if the sequence of its partial sums converges. Sum of convergent series is the limit of the sequence of its partial sums.
Definition. If the sequence of partial sums of a series diverges, i.e. has no limit, or has an infinite limit, then the series is called divergent and no amount is assigned to it.
Properties of rows.
1) The convergence or divergence of the series will not be violated if you change, discard or add a finite number of terms of the series.
2) Consider two rows 
And 
, where C is a constant number.
Theorem.
If the row 
converges and its sum is equalS, then the series 
also converges, and its sum is equal to CS.
(C
0)
3) Consider two rows 
And 
.Amount or difference of these series will be called a series 
, where the elements are obtained by adding (subtracting) the original elements with the same numbers.
Theorem.
If the rows 
And 
converge and their sums are equal respectivelySAnd
, then the series 
also converges and its sum is equalS
+
.
The difference of two convergent series will also be a convergent series.
The sum of a convergent and a divergent series is a divergent series.
It is impossible to make a general statement about the sum of two divergent series.
When studying series, they mainly solve two problems: studying convergence and finding the sum of the series.
Cauchy criterion.
(necessary and sufficient conditions for the convergence of the series)
In order for the sequence 
was convergent, it is necessary and sufficient that for any 
there was such a numberN, that atn
>
Nand anyp> 0, where p is an integer, the following inequality would hold:
.
Proof. (necessity)
Let 
, then for any number
there is a number N such that the inequality
is fulfilled when n>N. For n>N and any integer p>0 the inequality also holds 
. Taking into account both inequalities, we obtain:
The need has been proven. We will not consider the proof of sufficiency.
Let us formulate the Cauchy criterion for the series.
In order for the series 
was convergent, it is necessary and sufficient that for any 
there was a numberNsuch that atn>
Nand anyp>0 the inequality would hold
.
However, in practice, using the Cauchy criterion directly is not very convenient. Therefore, as a rule, simpler convergence tests are used:
1) If the row
converges, then it is necessary that the common term u n tended to zero. However, this condition is not sufficient. We can only say that if the common term does not tend to zero, then the series definitely diverges. For example, the so-called harmonic series 
is divergent, although its common term tends to zero.
Example. Investigate the convergence of the series 
We'll find 
- the necessary criterion for convergence is not satisfied, which means the series diverges.
2) If a series converges, then the sequence of its partial sums is bounded.
However, this sign is also not sufficient.
For example, the series 1-1+1-1+1-1+ … +(-1) n +1 +… diverges, because the sequence of its partial sums diverges due to the fact that
However, the sequence of partial sums is limited, because 
at any n.
Series with non-negative terms.
When studying series of constant sign, we will limit ourselves to considering series with non-negative terms, because simply multiplying by –1 from these series can yield series with negative terms.
Theorem.
For the convergence of the series 
with non-negative terms it is necessary and sufficient for the partial sums of the series to be bounded.
A sign for comparing series with non-negative terms.
Let two rows be given
And 
at u n ,
v n
0
.
Theorem.
If u n
v n at any n, then from the convergence of the series 
the series converges
, and from the divergence of the series
the series diverges
.
Proof.
Let us denote by S n
And
n partial sums of series
And 
. Because according to the conditions of the theorem, the series 
converges, then its partial sums are bounded, i.e. in front of everyone n n M, where M is a certain number. But because u n
v n, That S n
n then the partial sums of the series
are also limited, and this is sufficient for convergence.
Example. Examine the series for convergence 
Because 
, and the harmonic series 
diverges, then the series diverges 
.
Example.
Because 
, and the series 
converges (like a decreasing geometric progression), then the series 
also converges.
The following convergence sign is also used:
Theorem.
If 
and there is a limit 
, Whereh– a number other than zero, then the series 
And 
behave identically in terms of convergence.
D'Alembert's sign.
(Jean Leron d'Alembert (1717 - 1783) - French mathematician)
If for a series 
with positive terms there is such a numberq<1,
что для всех достаточно больших
ninequality holds
then a series 
converges, if for all there are sufficiently largencondition is met
then a series 
diverges.
D'Alembert's limiting sign.
D'Alembert's limiting criterion is a consequence of the above D'Alembert criterion.
If there is a limit 
, then when
< 1 ряд сходится, а при
> 1 – diverges. If
= 1, then the question of convergence cannot be answered.
Example. Determine the convergence of the series 
.
Conclusion: the series converges.
Example. Determine the convergence of the series 
Conclusion: the series converges.
Cauchy's sign. (radical sign)
If for a series 
with non-negative terms there is such a numberq<1,
что для всех достаточно больших
ninequality holds
,
then a series 
converges, if for all there are sufficiently largeninequality holds
then a series 
diverges.
Consequence.
If there is a limit 
, then when<1
ряд сходится, а при >Row 1 diverges.
Example. Determine the convergence of the series 
.
Conclusion: the series converges.
Example. Determine the convergence of the series 
.
Those. The Cauchy test does not answer the question of the convergence of the series. Let us check that the necessary convergence conditions are satisfied. As mentioned above, if a series converges, then the common term of the series tends to zero.
,
Thus, the necessary condition for convergence is not satisfied, which means the series diverges.
Integral Cauchy test.
If
(x) is a continuous positive function decreasing over the interval And 
then the integrals 
And 
behave identically in terms of convergence.
Alternating series.
Alternating rows.
An alternating series can be written as:
Where 
Leibniz's sign.
If the sign of the alternating row
absolute valuesu i are decreasing 
and the common term tends to zero 
, then the series converges.
Absolute and conditional convergence of series.
Let's consider some alternating series (with terms of arbitrary signs).
(1)
and a series composed of the absolute values of the members of the series (1):
(2)
Theorem. From the convergence of series (2) follows the convergence of series (1).
Proof. Series (2) is a series with non-negative terms. If series (2) converges, then by the Cauchy criterion for any >0 there is a number N such that for n>N and any integer p>0 the following inequality is true:
According to the property of absolute values:
That is, according to the Cauchy criterion, from the convergence of series (2) the convergence of series (1) follows.
Definition.
Row 
called absolutely convergent, if the series converges 
.
It is obvious that for series of constant sign the concepts of convergence and absolute convergence coincide.
Definition.
Row 
called conditionally convergent, if it converges and the series 
diverges.
D'Alembert's and Cauchy's tests for alternating series.
Let 
- alternating series.
D'Alembert's sign.
If there is a limit 
, then when<1
ряд
will be absolutely convergent, and when>
Cauchy's sign.
If there is a limit 
, then when<1
ряд
will be absolutely convergent, and if >1 the series will be divergent. When =1, the sign does not give an answer about the convergence of the series.
Properties of absolutely convergent series.
1)
Theorem.
For absolute convergence of the series 
it is necessary and sufficient that it can be represented as the difference of two convergent series with non-negative terms.
Consequence. A conditionally convergent series is the difference of two divergent series with non-negative terms tending to zero.
2) In a convergent series, any grouping of the terms of the series that does not change their order preserves the convergence and magnitude of the series.
3) If a series converges absolutely, then the series obtained from it by any permutation of terms also converges absolutely and has the same sum.
By rearranging the terms of a conditionally convergent series, one can obtain a conditionally convergent series having any predetermined sum, and even a divergent series.
4) Theorem. For any grouping of members of an absolutely convergent series (in this case, the number of groups can be either finite or infinite, and the number of members in a group can be either finite or infinite), a convergent series is obtained, the sum of which is equal to the sum of the original series.
5) If the rows 
And 
converge absolutely and their sums are equal respectively S
and , then a series composed of all products of the form 
taken in any order, also converges absolutely and its sum is equal to S
- the product of the sums of the multiplied series.
If you multiply conditionally convergent series, you can get a divergent series as a result.
Functional sequences.
Definition. If the members of the series are not numbers, but functions of X, then the series is called functional.
The study of the convergence of functional series is more complicated than the study of numerical series. The same functional series can, with the same variable values X converge, and with others - diverge. Therefore, the question of convergence of functional series comes down to determining those values of the variable X, at which the series converges.
The set of such values is called area of convergence.
Since the limit of each function included in the convergence region of the series is a certain number, the limit of the functional sequence will be a certain function:
Definition. Subsequence ( f n (x) } converges to function f(x) on the segment if for any number >0 and any point X from the segment under consideration there is a number N = N(, x), such that the inequality
is fulfilled when n>N.
With the selected value >0, each point of the segment has its own number and, therefore, there will be an infinite number of numbers corresponding to all points of the segment. If you choose the largest of all these numbers, then this number will be suitable for all points of the segment, i.e. will be common to all points.
Definition. Subsequence ( f n (x) } converges uniformly to function f(x) on the segment , if for any number >0 there is a number N = N() such that the inequality
is fulfilled for n>N for all points of the segment.
Example. Consider the sequence 
This sequence converges on the entire number line to the function f(x)=0 , because
Let's build graphs of this sequence:
sinx 
As can be seen, with increasing number n the sequence graph approaches the axis X.
Functional series.
Definition.
Private (partial) amounts functional range 
functions are called 
Definition.
Functional range 
called convergent at point ( x=x 0
), if the sequence of its partial sums converges at this point. Sequence limit 
called amount row 
at the point X 0
.
Definition.
Set of all values X, for which the series converges 
called area of convergence row.
Definition.
Row 
called uniformly convergent on the interval if the sequence of partial sums of this series converges uniformly on this interval.
Theorem. (Cauchy criterion for uniform convergence of series)
For uniform convergence of the series 
it is necessary and sufficient that for any number
>0 such a number existedN(
), which atn>
Nand any wholep>0 inequality
would hold for all x on the interval [a, b].
Theorem. (Weierstrass test for uniform convergence)
(Karl Theodor Wilhelm Weierstrass (1815 – 1897) – German mathematician)
Row 
converges uniformly and absolutely on the interval [a,
b], if the moduli of its terms on the same segment do not exceed the corresponding terms of a convergent number series with positive terms:
those. there is an inequality:
.
They also say that in this case the functional series 
is majorized number series 
.
Example. Examine the series for convergence 
.
Because 
always, it is obvious that 
.
Moreover, it is known that the general harmonic series 
when=3>1 converges, then, in accordance with the Weierstrass test, the series under study converges uniformly and, moreover, in any interval.
Example. Examine the series for convergence 
.
On the interval [-1,1] the inequality holds 
those. according to the Weierstrass criterion, the series under study converges on this segment, but diverges on the intervals (-, -1) (1, ).
Properties of uniformly convergent series.
1) Theorem on the continuity of the sum of a series.
If the members of the series 
- continuous on the segment [a,
b] function and the series converges uniformly, then its sumS(x) is a continuous function on the interval [a,
b].
2) Theorem on term-by-term integration of a series.
Uniformly converging on the segment [a, b] a series with continuous terms can be integrated term by term on this interval, i.e. a series composed of integrals of its terms over the segment [a, b] , converges to the integral of the sum of the series over this segment.
3) Theorem on term-by-term differentiation of a series.
If the members of the series 
converging on the segment [a,
b] represent continuous functions having continuous derivatives, and a series composed of these derivatives 
converges uniformly on this segment, then this series converges uniformly and can be differentiated term by term.
Based on the fact that the sum of the series is some function of the variable X, you can perform the operation of representing a function in the form of a series (expansion of a function into a series), which is widely used in integration, differentiation and other operations with functions.
In practice, power series expansion of functions is often used.
Power series.
Definition. Power series is called a series of the form
.
To study the convergence of power series, it is convenient to use D'Alembert's test.
Example. Examine the series for convergence 
We apply d'Alembert's sign:
.
We find that this series converges at 
and diverges at 
.
Now we determine the convergence at the boundary points 1 and –1.
For x = 1: 
The series converges according to Leibniz's criterion (see Leibniz's sign.).
At x = -1: 
the series diverges (harmonic series).
Abel's theorems.
(Nils Henrik Abel (1802 – 1829) – Norwegian mathematician)
Theorem.
If the power series 
converges atx
=
x 1
, then it converges and, moreover, for absolutely everyone 
.
Proof. According to the conditions of the theorem, since the terms of the series are limited, then
Where k- some constant number. The following inequality is true:
From this inequality it is clear that when x<
x 1
the numerical values of the terms of our series will be less (at least not more) than the corresponding terms of the series on the right side of the inequality written above, which form a geometric progression. The denominator of this progression 
according to the conditions of the theorem, it is less than one, therefore, this progression is a convergent series.
Therefore, based on the comparison criterion, we conclude that the series 
converges, which means the series 
converges absolutely.
Thus, if the power series 
converges at a point X 1
, then it converges absolutely at any point in the interval of length 2 
centered at a point X
= 0.
Consequence.
If at x = x 1
the series diverges, then it diverges for everyone 
.
Thus, for each power series there is a positive number R such that for all X such that 
the series is absolutely convergent, and for all 
the row diverges. In this case, the number R is called radius of convergence. The interval (-R, R) is called convergence interval.
Note that this interval can be closed on one or both sides, or not closed.
The radius of convergence can be found using the formula:
Example. Find the area of convergence of the series 
Finding the radius of convergence 
.
Therefore, this series converges for any value X. The common term of this series tends to zero.
Theorem.
If the power series 
converges for a positive value x=x 1
, then it converges uniformly in any interval inside 
.
Actions with power series.