Example. Find the sum of the series

$$
\sum_{n=1}^{\infty} \frac{\sin ^{n} x}{n}
$$

and specify the domain 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 $|\sin x|<1$.

At the boundary points, when $x=\pi / 2+2 \pi k$ the series diverges, and when $x=3 \pi / 2+2 \pi k$ the series converges conditionally.

Therefore, the given series converges for all $x \neq \pi / 2+2 \pi k$ $(k=0, \pm 1, \ldots)$.

2. Make the substitution $\sin x=t$. We obtain a geometric series (1) with the interval of convergence $[-1,1)$.
3. Use the formula for the sum of the terms of an infinitely decreasing geometric progression

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

4. Moreover, we have the obvious equality

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

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 (4), we get

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

Note that since the series (1) converges at the boundary point $t=-1$, its sum is continuous at this point (from the right). Therefore, formula (5) is valid for all $t \in[-1,1)$.

6. Replacing $t$ with $\sin x$, we obtain for $x \neq \pi / 2+2 \pi k$

$$
S(x)=-\ln (1-\sin x) .
$$

Answer. $\quad S(x)=-\ln (1-\sin x), \quad x \neq \pi / 2+2 \pi k$.

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

1. $\sum_{n=1}^{\infty} \frac{6^{n} x^{n}}{n}$.
2. $\sum_{n=0}^{\infty} \frac{\left(1-16 x^{4}\right)^{n}}{n+1}$.
3. $\sum_{n=0}^{\infty} \frac{2^{n}}{(n+1) x^{2 n}}$.
4. $\sum_{n=1}^{\infty} \frac{3^{n}}{n(n+1) x^{n}}$.
5. $\sum_{n=1}^{\infty} \frac{(-1)^{n} 4^{n} x^{2 n}}{2 n+1}$.
6. $\sum_{n=1}^{\infty} \frac{x^{n+2}}{n(n+1)}$.
7. $\sum_{n=1}^{\infty} \frac{3^{n-1}}{n x^{n-1}}$.
8. $\sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}}{n(n+1)}$.
9. $\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{2 n+1}$.
10. $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2 n}}{n(2 n-1)}$.

Answers.

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

### 10.12. Summation of a Series by Term-by-Term Differentiation

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

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

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

Plan of Solution.

1. Find the interval of convergence of the series.

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

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

If $f(x)= \pm 1$, the series diverges (the necessary condition for convergence is not satisfied). Therefore, the interval of convergence is determined by the inequalities $-1<f(x)<1$.

2. In the original series, make the substitution $f(x)=t$ and write it as the sum of two series

$$
\sum_{n=k}^{\infty} n t^{n}+b \sum_{n=k}^{\infty} t^{n}
$$

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

$$
\sum_{n=k}^{\infty} t^{n} \quad \text { and } \quad \sum_{n=k}^{\infty} n t^{n}
$$

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. Moreover, we have the obvious equality

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

5. Considering that a power series can be differentiated term by term at any point in the interval of convergence, and using formula (1), we get

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

6. Compute the derivative and substitute $t$ with $f(x)$. Write the answer: the sum of the series and the interval of its convergence.

Note. If the series has the form

$$
\sum_{n=k}^{\infty}\left(n^{2}+b n+c\right) f(x)^{n}
$$

then we compute the sum of three series, and when computing the sum of the series

$$
\sum_{n=k}^{\infty} n^{2} f(x)^{n}
$$

we apply the theorem on term-by-term differentiation of a power series twice.