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. Featured below online calculator, based on the Wolfram Alpha system, allows you to calculate the sum of fairly 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).

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 compose nth partial the 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. The issue of decomposition rational fractions a separate topic is devoted to the 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 parts 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. 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 the standard course higher mathematics usually they are content with “crossing out” the canceling terms, without requiring any evidence. So we have an expression for nth partial sums: $S_n=\frac(1)(3)-\frac(1)(2n+3)$. 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 $\frac(1)(2k+3)$. Final expression for a partial amount 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 is infinite number sequence 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.

Number series. Convergence and divergence of number series. D'Alembert's convergence test. Alternating series. Absolute and conditional convergence of series. Functional series. Power series. Decomposition elementary functions in the Maclaurin series.

Guidelines on topic 1.4:

Number series:

A number series is a sum of the form

where are the numbers u 1, u 2, u 3, n n, called members of a series, form an infinite sequence; the term un is called the common term of the series.

. . . . . . . . .

composed of the first terms of the series (27.1) are called partial sums of this series.

Each row can be associated with a sequence of partial sums S 1, S 2, S 3. If, with an infinite increase in number n, the partial sum of the series S n tends to the limit S, then the series is called convergent, and the number S- the sum of a convergent series, i.e.

This entry is equivalent to

If the partial amount S n series (27.1) with unlimited increase n does not have a finite limit (in particular, it tends to + ¥ or to - ¥), then such a series is called divergent

If the series converges, then the value S n for a sufficiently large n is an approximate expression for the sum of the series S.

Difference r n = S - S n is called the remainder of the series. If the series converges, then its remainder tends to zero, i.e. r n = 0, and vice versa, if the remainder tends to zero, then the series converges.

A series of the form is called geometric series.

called harmonic.

If N®¥, then S n®¥, i.e. the harmonic series diverges.

Example 1. Write a series based on its given common term:

1) putting n = 1, n = 2, n = 3, we have an infinite sequence of numbers: , , , Adding its terms, we get the series

2) Doing the same, we get the series

3) Giving n the values ​​1, 2, 3, and considering that 1! = 1, 2! = 1 × 2.3! = 1 × 2 × 3, we get the series

Example 2. Find n-th member of the series according to its given first numbers:

1) ; 2) ; 3) .

Example 3. Find the sum of the terms of the series:

2) .

1) Find the partial sums of the terms of the series:

; ;

… .

Let's write down the sequence of partial sums: …, , … .

The common term of this sequence is . Hence,

.

The sequence of partial sums has a limit equal to . So, the series converges and its sum is equal to .

2) It is infinitely decreasing geometric progression, in which a 1 = , q= . Using the formula, we get: This means that the series converges and its sum is equal to 1.

Convergence and divergence of number series. Sign of convergence d'Alembert :

A necessary sign of convergence of a series. A series can converge only if its common term is u n with unlimited number increase n tends to zero:

If , then the series diverges - this is a sufficient sign of the solubility of the series.

Sufficient signs of convergence of a series with positive terms.

A sign for comparing series with positive terms. The series under study converges if its terms do not exceed the corresponding terms of another, obviously convergent series; the series under study diverges if its members exceed the corresponding members of another obviously divergent series.

When studying series for convergence and solubility based on this criterion, a geometric series is often used

which converges at |q|

,

being divergent.

When studying series, the generalized harmonic series is also used

.

If p= 1, then this series turns into a harmonic series, which is divergent.

If p< 1, то члены данного ряда больше соответствующих членов гармонического ряда и, значит, он расходится. При p> 1 we have a geometric series in which | q| < 1; он является сходящимся. Итак, обобщенный гармонический ряд сходится при p> 1 and diverges at p£1.

D'Alembert's sign. If for a series with positive terms

(u n >0)

the condition is satisfied, then the series converges at l l > 1.

D'Alembert's sign does not give an answer if l= 1. In this case, other techniques are used to study the series.

Alternating series.

Absolute and conditional convergence of series:

Number series

u 1 + u 2 + u 3 + u n

is called alternating if among its members there are both positive and negative numbers.

A number series is called alternating if any two adjacent terms have opposite signs. This series is a special case of an alternating series.

Convergence test for alternating series. If the terms of an alternating series monotonically decrease in absolute value and common member u n tends to zero as n® , then the series converges.

A series is said to be absolutely convergent if the series also converges. If a series converges absolutely, then it is convergent (in the usual sense). The reverse statement is not true. A series is called conditionally convergent if it itself converges, and the series composed of the moduli of its members diverges. Example 4. Examine the series for convergence .

Let us apply Leibniz's sufficient test for alternating series. We get because the . Therefore, this series converges. Example 5. Examine the series for convergence .

Let's try to apply Leibniz's criterion: It can be seen that the modulus of the general term does not tend to zero when n → ∞. Therefore, this series diverges. Example 6. Determine whether a series is absolutely convergent, conditionally convergent, or divergent.

Applying d'Alembert's test to a series composed of the modules of the corresponding terms, we find Therefore, this series converges absolutely.

Example 7. Examine the sign-alternating series for convergence (absolute or conditional):

1) The terms of this series monotonically decrease in absolute value And . Therefore, according to Leibniz's criterion, the series converges. Let us find out whether this series converges absolutely or conditionally.

2) The terms of this series monotonically decrease in absolute value: , But

.

Functional rows:

A regular number series consists of numbers:

All members of the series - This numbers.

The functional series consists of functions:

In addition to polynomials, factorials, etc., the general term of the series certainly the letter "x" is included. For example, it looks like this: . Like a number series, any functional series can be written in expanded form:

As you can see, all members of the functional series are functions.

The most popular type of functional series is power series.

Power series:

Power series is called a series of the form

,

where are the numbers a 0, a 1, a 2, a n are called coefficients of the series, and the term a n x n- a common member of the series.

Area of ​​convergence power series is called the set of all values x, for which this series converges.

Number R is called the radius of convergence of the series if at | x| the series converges.

Example 8. Given a series

Investigate its convergence at points x= 1 and X= 3, x= -2.

When x = 1, this series turns into a number series

.

Let us investigate the convergence of this series using D'Alembert's criterion. We have

those. the series converges.

For x = 3 we get the series

Which diverges because the necessary criterion for the convergence of a series is not satisfied

For x = -2 we get

This is an alternating series, which, according to Leibniz's criterion, converges.

So, at points x= 1 and X= -2. the series converges, and at the point x= 3 diverges.

Expansion of elementary functions in Maclaurin series:

Near Taylor for function f(x) is called a power series of the form

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:

1. “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).
2. “Nper” is the number of periods for interest payments. In our example – 4 years.
3. "Plt" - periodic payments. In our case there are none. Therefore, we do not fill in the argument field.
4. “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 build 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.