Example 5.17. Find the interval of convergence of the power series

$$
1-\frac{x}{2 \cdot 2}+\frac{x^{2}}{3 \cdot 2^{2}}-\frac{x^{3}}{4 \cdot 2^{3}}+\ldots+(-1)^{n} \frac{x^{n}}{(n+1) 2^{n}}+\ldots
$$
Solution. Let's form a series of the absolute values of the terms of the given series:

$$
1+\frac{|x|}{2 \cdot 2}+\frac{|x|^{2}}{3 \cdot 2^{2}}+\frac{|x|^{3}}{4 \cdot 2^{3}}+\ldots+\frac{|x|^{n}}{(n+1) 2^{n}}+\ldots
$$

According to the D'Alembert's ratio test, the resulting positive series converges (absolutely) for those values of \( x \) for which \( l=\lim _{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}}<1 \). Here, \( u_{n}=\frac{|x|^{n}}{(n+1) 2^{n}}, \quad u_{n+1}=\frac{|x|^{n+1}}{(n+2) 2^{n+1}} \).

From this,

$$
l=\lim _{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}}=\lim _{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+2) 2^{n+1}}: \frac{|x|^{n}}{(n+1) 2^{n}}=\frac{|x|}{2} \lim _{n \rightarrow \infty} \frac{n+1}{n+2}=\frac{|x|}{2} .
$$

Determine for which values of \( x \) this limit \( l \) will be less than one. For this, solve the inequality \( \frac{|x|}{2}<1 \), or \( |x|<2 \), from which \( -2<x<2 \).

Thus, the original series converges (absolutely) in the interval \( (-2 ; 2) \) - this is the interval of convergence of the given series.

Let's investigate the convergence of the series at the endpoints of the interval of convergence. When \( x=-2 \), we obtain the numerical series

$$
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}+\ldots
$$

This is the harmonic series, which, as is known, diverges.

When \( x=2 \), we obtain the alternating numerical series

$$
1-\frac{1}{2}+\frac{1}{3}-\ldots+(-1)^{n} \frac{1}{n+1}+\ldots
$$

which converges (conditionally) by the Leibniz criterion.

Thus, the domain of convergence of the given series is \( -2<x \leq 2 \).