Example 4.26. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{n^{2} 2^{n}}$.
Solution. In this case, $a_{n}=\frac{1}{n^{2} 2^{n}}$. Using the ratio test (Theorem 4.23), we find $R$:

$$
R=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2} 2^{n}}}{\frac{1}{(n+1)^{2} 2^{n+1}}}=\lim _{n \rightarrow \infty} \frac{2(n+1)^{2}}{n^{2}}=2
$$

The series under investigation will converge in the interval determined by the inequality $|x-2|<2$ or $0<x<4$. Additional investigation is required at the points 0 and 4. When $x=0$, we obtain the alternating series $\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}}$, which converges absolutely. When $x=4$, we obtain the series of reciprocals of squares $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$, which also converges. The series under investigation converges absolutely for $x \in[0,4]=D$ and diverges for $x \notin D$.