Let's simplify the following sum:

$$
\frac{1}{2}+\frac{1}{3 \cdot 1!}+\frac{1}{4 \cdot 2!}+\ldots+\frac{1}{(n+1) \cdot(n-1)!}
$$
Some of the submitters achieved this by extending the terms of the sum with the missing factor. This perhaps gave the sum a somewhat more pleasing form, but it is not any easier to calculate for some $n$, and no other advantage is apparent. This extension is a good starting point in every case.

I. Solution: $\frac{1}{(k+1) \cdot(k-1)!}=\frac{k}{(k+1)!}=\frac{k+1-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$. Performing this transformation in every term of the sum:

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

Tamás Kővári (Bp.-i Evangelical Gymnasium VIII. grade)

II. Solution: Let's denote the sum in question by $S_{n}$. We will show by complete induction that $S_{n}=1-\frac{1}{(n+1)!}$.

$$
n=1 \text { is true: } S_{1}=\frac{1}{2}=1-\frac{1}{2!}
$$

Now we will show that if the statement is true for some $n$, then it also follows that it is true for the next number, $n+1$. It is as if it is inherited from every integer to the next, and thus it is true for every $n$.

Indeed, if $S_{n}=1-\frac{1}{(n+1)!}$, then

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

Solved by: Gehér L., Gősy S., Vermes R.