E x a m p l e 1. Let's find the limit of the sum

$$
S_{n}=\frac{3}{4}+\frac{5}{36}+\ldots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$

as $n \rightarrow \infty$.
S o l u t i o n. First, let's try to simplify $S_{n}$. Since the fraction 56
$\frac{2 k+1}{k^{2}(k+1)^{2}}$ can be represented as the difference $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}$, then

$$
\begin{gathered}
\frac{3}{4}=1-\frac{1}{2^{2}} \\
\frac{5}{36}=\frac{1}{2^{2}}-\frac{1}{3^{2}} \\
\frac{2 n+1}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}
\end{gathered}
$$

By adding these equalities, we get:

$$
S_{n}=1-\frac{1}{(n+1)^{2}}
$$

Then

$$
\lim _{n \rightarrow \infty} S_{n}=\lim _{n \rightarrow \infty}\left(1-\frac{1}{(n+1)^{2}}\right)=1
$$