Example. Find the sum of the series

$$
\sum_{n=0}^{\infty}(n+6) x^{7 n}
$$

and specify the interval of convergence of the series to this sum.
## Solution.

1. Find the interval of convergence of the series.

By the Cauchy criterion, the interval of convergence is determined by the inequality $\left|x^{7}\right|<1$. Hence, $-1<x<1$. At the boundary points $x= \pm 1$, the series diverges because the necessary condition for convergence is not satisfied. Therefore, the series converges in the interval ( $-1,1$ ).

2. Make the substitution $x^{7}=t$ in the original series and write it as the sum of two series

$$
S(t)=6 \sum_{n=0}^{\infty} t^{n}+\sum_{n=0}^{\infty} n t^{n}=6 S_{1}(t)+S_{2}(t)
$$

Thus, it is sufficient to find the sums of the series

$$
S_{1}(t)=\sum_{n=0}^{\infty} t^{n}, \quad S_{2}(t)=\sum_{n=1}^{\infty} n t^{n}
$$

3. Use the formula for the sum of the terms of an infinite decreasing geometric progression:

$$
\sum_{n=0}^{\infty} t^{n}=\frac{1}{1-t}, \quad|t|<1
$$

Thus, $S_{1}(t)=\frac{1}{1-t}$ for all $t \in(-1,1)$.

4. Moreover, we have the obvious equality

$$
\sum_{n=1}^{\infty} n t^{n}=t \sum_{n=1}^{\infty} n t^{n-1}=t \sum_{n=1}^{\infty} \frac{d}{d t} t^{n}
$$

5. Considering that a power series can be differentiated term by term at any point in the interval of convergence, and using formula (2), we get

$$
S_{2}(t)=\sum_{n=1}^{\infty} n t^{n}=t \frac{d}{d t} \sum_{n=1}^{\infty} t^{n}=t \frac{d}{d t} \frac{t}{1-t}=\frac{t}{(1-t)^{2}}, \quad t \in(-1,1)
$$

Thus,

$$
S(t)=6 S_{1}(t)+S_{2}(t)=\frac{6}{1-t}+\frac{t}{(1-t)^{2}}=\frac{6-5 t}{(1-t)^{2}}, \quad t \in(-1,1)
$$

Substituting $t$ with $x^{7}$, we get

$$
\sum_{n=0}^{\infty}(n+6) x^{7 n}=\frac{6-5 x^{7}}{\left(1-x^{7}\right)^{2}} \quad \forall x \in(-1,1)
$$

Answer. $S(x)=\frac{6-5 x^{7}}{\left(1-x^{7}\right)^{2}}, \quad x \in(-1,1)$.

Conditions of the Problems. Find the sums of the functional series and specify the intervals of their convergence to these sums.

1. $\sum_{n=1}^{\infty} n x^{n+1}$.
2. $\quad \sum_{n=1}^{\infty} n 2^{n} x^{n}$.
3. $\sum_{n=0}^{\infty}(n+1) x^{2 n+4}$.
4. $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} n x^{n-1}}{3^{n-1}}$.
5. $\quad \sum_{n=0}^{\infty} \frac{(n+1)(n+2) x^{n}}{2^{n}}$.
6. $\sum_{n=1}^{\infty} n^{2} x^{n-1}$.
7. $\quad \sum_{n=0}^{\infty}\left(2 n^{2}-2 n+1\right) x^{n}$.
8. $\sum_{n=1}^{\infty}(-1)^{n-1} n(n+1) x^{n}$.
9. $\quad \sum_{n=1}^{\infty} n(n+2) x^{n}$.
10. $\sum_{n=1}^{\infty} n\left(x^{3}+1\right)^{n-1}$.

Answers.

1. $S=\frac{x^{2}}{(1-x)^{2}}, \quad x \in(-1,1)$.
2. $S=\frac{2 x}{(1-2 x)^{2}}, x \in\left(-\frac{1}{2}, \frac{1}{2}\right)$.
3. $S=\frac{x^{4}}{\left(1-x^{2}\right)^{2}}, \quad x \in(-1,1)$.
4. $S=\frac{9}{(x+3)^{2}}, \quad x \in(-3,3)$.
5. $S=\frac{16}{(2-x)^{3}}, \quad x \in(-2,2)$.
6. $S=\frac{1+x}{(1-x)^{3}}, \quad x \in(-1,1)$.
7. $S=\frac{5 x^{2}-2 x+1}{(1-x)^{3}}, x \in(-1,1)$.
8. $S=\frac{2 x}{(1+x)^{3}}, \quad x \in(-1,1)$.
9. $S=\frac{3 x-x^{2}}{(1-x)^{3}}, \quad x \in(-1,1)$.
10. $S=\frac{1}{x^{6}}, \quad x \in(-\sqrt[3]{2}, 0)$.

### 10.13. Taylor Series

Problem Statement. Expand the function $f(x)$ in a Taylor series in powers of $x\left(x_{0}=0\right)$.

## Plan of Solution.

1. Transform the function $f(x)$ into a form that allows the use of tabulated expansions $e^{x}, \sin x, \cos x,(1+x)^{m}, \ln (1+x)$.
2. Find the Taylor series expansion of the function using tabulated expansions, addition (subtraction) of series, and multiplication of a series by a number.
3. Determine the interval of convergence of the obtained series to the function $f(x)$.

Remark. If it is necessary to expand the function in a Taylor series in powers of $x-x_{0}$, first make the substitution $t=x-x_{0}$, find the expansion in powers of $t$, and then return to the variable $x$.