Example 5. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{x^{2 n-1}}{\left(n^{2}+1\right) \cdot 3^{n}}$.
Solution. For this series, the formulas of D'Alembert and Cauchy are not applicable - the corresponding limits do not exist (the coefficients of the series at even powers of $x$ are zero, while the coefficients at odd powers are non-zero; thus, in particular, $\frac{a_{2 n}}{a_{2 n+1}}=0$, and $\frac{a_{2 n+1}}{a_{2 n+2}}=\infty$).

We proceed as follows. Assume that $x=x_{0}$ is a fixed number, and find the conditions under which the sign-positive series $\sum_{n=1}^{\infty} \frac{\left|x_{0}\right|^{2 n-1}}{\left(n^{2}+1\right) \cdot 3^{n}}$ converges. For this, we apply Theorem 7 with $a_{n}\left(x_{0}\right)=\frac{\left|x_{0}\right|^{2 n-1}}{\left(n^{2}+1\right) 3^{n}}$:

$$
q=\lim _{n \rightarrow \infty} \frac{a_{n+1}\left(x_{0}\right)}{a_{n}\left(x_{0}\right)}=\lim _{n \rightarrow \infty} \frac{\left|x_{0}\right|^{2 n+1}}{\left((n+1)^{2}+1\right) \cdot 3^{n+1}} \cdot \frac{\left(n^{2}+1\right) \cdot 3^{n}}{\left|x_{0}\right|^{2 n-1}}=\frac{\left|x_{0}\right|^{2}}{3}
$$

The condition for the convergence of the series is given by the inequality $q < 1$, which simplifies to $\left|x_{0}\right| < \sqrt{3}$. At $x_{0} = \pm \sqrt{3}$, we obtain the convergent series $\pm \sum_{n=1}^{\infty} \frac{1}{\sqrt{3}\left(n^{2}+1\right)}$.

Answer. The series converges absolutely for $-\sqrt{3} \leqslant x \leqslant \sqrt{3}$ and diverges outside this interval.

## Exercises

Find the intervals of convergence for the series.

1. $\sum_{n=2}^{\infty} \frac{x^{n}}{n \ln ^{2} n}$. 2. $\sum_{n=1}^{\infty} \frac{x^{n}}{n^{n}}$. 3. $\sum_{n=0}^{\infty} n!x^{n}$. 4. $\sum_{n=2}^{\infty} \frac{x^{n}}{\ln n}$.
2. $\sum_{n=0}^{\infty} \frac{1}{\ln ^{2}\left(n^{2}+1\right)}\left(\frac{2 x-1}{x+2}\right)^{n}$. 6. $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1) \cdot 4^{n}(x+2)^{n}}$.
3. $\sum_{n=0}^{\infty}\left(\frac{n}{2 n+1}\right)^{n} \cdot x^{n}$. 8. $\sum_{n=2}^{\infty} \frac{x^{n-1}}{2^{n} \cdot n \ln 2 n}$. 9. $\sum_{n=0}^{\infty} \frac{n^{2}}{n+1} \cdot x^{2 n}$.
4. $\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{n}(x+1)^{n}$. 11. $\sum_{n=0}^{\infty} \frac{n}{n+1} \cdot\left(\frac{x}{2}\right)^{n}$.
5. $\sum_{n=2}^{\infty}(-1)^{n} \frac{(x-3)^{n}}{\sqrt[3]{n} \sqrt{n-1}}$.

## Answers

1. $[-1 ; 1]$. 2. $(-\infty ;+\infty)$. 3. $\{0\}$. 4. $[-1 ; 1)$. 5. $\left(-\frac{1}{3} ; 3\right)$.
2. $\left(-\infty ;-\frac{9}{4}\right) \cup\left[-\frac{7}{4} ;+\infty\right)$. 7. $(-2 ; 2)$. 8. $[-2 ; 2)$. 9. $(-1 ; 1)$. 10. $(-2 ; 0)$.
3. $(-2 ; 2)$. 12. $(2 ; 4]$.

## §4. Application of Series in Approximate Calculations. Expansion of Functions into Power Series

$1^{\circ}$. A function $\hat{f}(x)$ is called infinitely differentiable in some interval $(a, b)$ if in this interval it has continuous derivatives of any order $n \in N$.

Theorem 18 (Taylor-Maclaurin). Suppose that the function $f(x)$ is infinitely differentiable in some interval $\left(x_{0}-l ; x_{0}+l\right)$ and all its derivatives are bounded by the same number $M$ in this interval:

$$
\left|f^{(n)}(x)\right| \leqslant M, \quad x \in\left(x_{0}-l ; x_{0}+l\right)
$$

Then in this interval $f(x)$ can be represented as the sum of a generalized power series with center at the point $x_{0}$:

$$
\begin{gathered}
f(x)=c_{0}+c_{1}\left(x-x_{0}\right)+c_{2}\left(x-x_{0}\right)^{2}+\ldots+c_{n}\left(x-x_{0}\right)^{n}+\ldots \\
x \in\left(x_{0}-l ; x_{0}+l\right)
\end{gathered}
$$

where $c_{0}=f\left(x_{0}\right), c_{1}=f^{\prime}\left(x_{0}\right), \ldots, c_{n}=\frac{f^{(n)}\left(x_{0}\right)}{n!}, n=2,3, \ldots$

When $x_{0}=0$, the corresponding power series and its coefficients belong to Maclaurin and bear his name. The Maclaurin series is the most widely used in practical calculations involving elementary functions.

$2^{\circ}$. The following expansions into Maclaurin series are verified directly (these expansions are easy to remember):

(1) $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{n}}{n!}+\ldots, \quad x \in(-\infty ;+\infty)$;

(2) $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\ldots+(-1)^{n} \frac{x^{2 n}}{(2 n)!}+\ldots, x \in(-\infty ;+\infty)$;

(3) $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\ldots+(-1)^{n-1} \frac{x^{2 n-1}}{(2 n-1)!}+\ldots$, $x \in(-\infty ;+\infty)$

(4) $(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!} x^{2}+\ldots+$

$$
+\frac{\alpha(\alpha-1)(\alpha-2) \cdot \ldots \cdot(\alpha-n+1)}{n!}+a^{n} \ldots, x \in(-1 ; 1)
$$

(5) $\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots+(-1)^{n-1} \frac{x^{n}}{n}+\ldots, x \in(-1 ; 1)$.

Series (4) is called the binomial series, and series (5) is called the logarithmic series.

$3^{\circ}$. Using arithmetic operations, differentiation, integration, substitution of variables, and other operations, we can obtain Maclaurin series expansions for a large number of elementary functions from the given series.

The obtained series converge in the intersection of the intervals of convergence of the series involved in the corresponding intermediate operations.

For example, the Maclaurin series for the function $\cos x-\sin 2 x$ can be obtained as follows: from series (2) we subtract series (3), in which we first substitute $x$ with $2 x$:

$$
\begin{aligned}
& \text { substitute } x \text { with 2x: } \\
& \begin{aligned}
\cos x-\sin 2 x=1-2 x-\frac{x^{2}}{2!}+\frac{2^{3} \cdot x^{3}}{3!}+\frac{x^{4}}{4!}- & \frac{2^{5} \cdot x^{5}}{5!}-\frac{x^{6}}{6!}+\frac{2^{7} \cdot x^{7}}{7!}+ \\
& +\frac{x^{8}}{8!}-\ldots, x \in(-\infty ;+\infty)
\end{aligned}
\end{aligned}
$$

## Examples with Solutions