Example 2. Find the interval of convergence of the series

$$
\sum_{n=1}^{\infty} \frac{1}{n+1}(x-2)^{2 n}
$$
Solution.

1. In this case, $c_{n}=0$ for all odd $n$. Therefore, we cannot use the formulas for the radius of convergence of a power series.

We use the Cauchy criterion, for which we compute

$$
\rho(x)=\lim _{n \rightarrow \infty} \sqrt[n]{\left|f_{n}(x)\right|}
$$

where $f_{n}(x)=\frac{1}{n+1}(x-2)^{2 n}$.

We find $\rho(x)$:

$$
\rho(x)=\lim _{n \rightarrow \infty} \sqrt[n]{\frac{1}{n+1}(x-2)^{2 n}}=(x-2)^{2}
$$

By the Cauchy criterion, the series converges when $\rho(x) < 1$ and diverges when $\rho(x) > 1$. Therefore, the interval of convergence is determined by the inequality $(x-2)^{2}<1$ and is $(1,3)$.

2. We investigate the convergence of the series at the boundary points of the interval of convergence. At the point $x=1$, the series

$$
\sum_{n=1}^{\infty} \frac{1}{n+1}(-1)^{n}
$$

converges conditionally by the Leibniz criterion.

At the point $x=3$, the series

$$
\sum_{n=1}^{\infty} \frac{1}{n+1}(1)^{n}
$$

diverges.

Answer. The domain of convergence of the power series is $[1,3)$.

Conditions of the Problem. Find the domains of convergence of the power series.

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

Answers. 1. $(-7,11)$. 2. $(2,4)$. 3. $(-7,-3)$. 4. $[-2,1)$. 5. $[-1,5)$. 6. $[1,5]$.
7. $[-4,-2]$.
8. $[-5,1)$.
9. $(-6,-2]$.
10. $(-\infty,+\infty)$.

### 10.11. Calculation of the Sum of a Series by Term-by-Term Integration

Problem Statement. Find the sum of a functional series of the form

$$
\sum_{n=k}^{\infty} \frac{f(x)^{n}}{n}
$$

and specify the domain of convergence of the series to this sum.

## PLAN OF SOLUTION.

1. Find the domain of convergence of the series.

By the Cauchy criterion, the interval of convergence is determined by the inequality

$$
|f(x)|<1
$$

If $f(x)=1$, the series diverges. If $f(x)=-1$, the series converges conditionally (by the Leibniz criterion). Therefore, the domain of convergence is determined by the inequalities $-1 \leq f(x)<1$.

2. Make the substitution $f(x)=t$ in the original series, resulting in the power series

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

with the domain of convergence $[-1,1)$.

3. The formula for the sum of the terms of an infinitely decreasing geometric progression is known:

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

4. Additionally, we have the obvious equality

$$
\sum_{n=k}^{\infty} \frac{t^{n}}{n}=\sum_{n=k}^{\infty} \int_{0}^{t} u^{n-1} d u, \quad t \in(-1,1)
$$

5. Considering that a power series can be integrated term by term on any interval $[0, t]$ entirely within the interval of convergence, and using formula (2), we obtain

$$
S(t)=\sum_{n=k}^{\infty} \frac{t^{n}}{n}=\int_{0}^{t} \sum_{n=k}^{\infty} u^{n-1} d u=\int_{0}^{t} \frac{u^{k}}{1-u} d u, \quad t \in(-1,1)
$$

Note that since the series (1) converges at the boundary point $t=-1$, the sum of the series is continuous at this point (from the right). Therefore, $S(-1)=\lim _{t \rightarrow-1+0} S(t)$.

6. Compute the integral, substitute $t$ with $f(x)$, and write the answer: the sum of the series and its domain of convergence.

Remark. If the series has the form

$$
\sum_{n=k}^{\infty} \frac{f(x)^{n}}{(n+a)(n+b)}
$$

then we apply the theorem on term-by-term integration of a power series twice or decompose the fraction into partial fractions:

$$
\frac{1}{(n+a)(n+b)}=\left(\frac{1}{n+a}-\frac{1}{n+b}\right) \frac{1}{b-a}
$$

and compute the sum of each series by term-by-term integration.