Example 4.1. Investigate the convergence of the series:

$$
\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\ldots+\frac{1}{4 n^{2}-1}+\ldots
$$
Solution. Let's transform the general term of this series:

$$
a_{n}=\frac{1}{4 n^{2}-1}=\frac{1}{(2 n-1)(2 n+1)}=\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)
$$

After this, it is not difficult to calculate the partial sums of the series:

$$
S_{n}=\sum_{p=1}^{n} a_{p}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2 n+1}\right)
$$

Since $\lim _{n \rightarrow \infty} S_{n}=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1-\frac{1}{2 n+1}\right)=\frac{1}{2}$, according to the definition, the series converges and its sum $S=\frac{1}{2}$, in other words,

$$
\sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1}=\frac{1}{2}
$$