Example 4.10. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{2 n+1}{a^{n}}$, $(a>0)$.
Solution. In this case $a_{n}=\frac{2 n+1}{a^{n}}>0$. Let's find the limit (4.6)

$$
\lim _{n \rightarrow \infty} \frac{2 n+3}{a^{n+1}}: \frac{2 n+1}{a^{n}}=\lim _{n \rightarrow \infty} \frac{2 n+3}{2 n+1} \cdot \frac{1}{a}=\frac{1}{a}
$$

Based on Theorem 4.8, we conclude: if $a>1$ the series converges, if $0<a<1$ the series diverges. When $a=1$ the series also diverges since its general term $a_{n}=2 n+1$ tends to infinity.