13.1. Calculate the sum

$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{n-1}{n!}
$$
13.1. We will prove by induction on $n$ that the considered sum equals $1-\frac{1}{n!}$. For $n=2$, we get the obvious equality $\frac{1}{2!}=1-\frac{1}{2!}$. Suppose that the required equality is proved for some $n$. Then

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