Example 1. Investigate the convergence of the functional series $\sum_{n=1}^{\infty} \frac{1}{n^{2}+3} \cdot\left(\frac{x+1}{x-1}\right)^{n}$
Solution. Let $q=q(x)=\frac{x+1}{x-1}$. The series $\sum_{n=1}^{\infty} \frac{q^{n}}{n^{2}+3}$ converges absolutely when $|q| \leqslant 1$ by the comparison theorem $5 .\left(\left|\frac{q^{n}}{n^{2}+3}\right| \leqslant \frac{1}{n^{2}+3}\right.$, if $|q|1$, then the series $\sum_{n=1}^{\infty} \frac{q^{n}}{n^{2}+3}$ diverges, because $\lim _{n \rightarrow \infty} \frac{q^{n}}{n^{2}+3} \neq 0$ when $|q|>1$ (prove this!). Thus, the given functional series converges absolutely under the condition $|q|=\left|\frac{x+1}{x-1}\right| \leqslant 1$. This is equivalent to the condition $x \leqslant 0$.

Answer. The given series converges absolutely on the ray ( $-\infty ; 0]$.