Task B-4.4. (20 points) Find the formula for the sum

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

Show that the sequence of sums converges and determine its limit!
Solution. First, rearrange the $n$-th term:

$$
\begin{aligned}
\frac{n+2}{n!+(n+1)!+(n+2)!} & =\frac{n+2}{n![1+(n+1)+(n+1)(n+2)]} \\
& =\frac{n+2}{n!\left(n^{2}+4 n+4\right)}=\frac{n+2}{n!(n+2)^{2}}=\frac{1}{n!(n+2)} \cdot \frac{n+1}{n+1} \\
& =\frac{n+1}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}
\end{aligned}
$$

Now we have

$$
\begin{aligned}
S_{n} & =\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\ldots+\frac{1}{n!}-\frac{1}{(n+1)!}+\frac{1}{(n+1)!}-\frac{1}{(n+2)!} \\
& =\frac{1}{2}-\frac{1}{(n+2)!}
\end{aligned}
$$

The limit of the sequence $S_{n}$ is $\lim _{n \rightarrow \infty}\left(\frac{1}{2}-\frac{1}{(n+2)!}\right)=\frac{1}{2}$.