For any $n$,

$$
a_{n}=\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\ldots+\frac{1}{n(n+1)(n+2)}
$$

Is the sequence $a_{n}$ convergent, and if so, what is its limit?
It seems useful to convert the product $\frac{1}{n(n+1)(n+2)}$ into a sum, as we hope that the sum of certain terms will be 0 and $a_{n}$ can be expressed more simply. Let's try to write $\frac{1}{n(n+1)(n+2)}$ in the form $\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}$:

$$
\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}=\frac{n^{2}(A+B+C)+n(3 A+2 B+C)+2 A}{n(n+1)(n+2)}
$$

If this holds for all $n$, then clearly

$$
\begin{array}{r}
A+B+C=0 \\
3 A+2 B+C=0 \\
2 A=1
\end{array}
$$

Solving this system of equations, we get $A=\frac{1}{2} ; B=-1 ; C=\frac{1}{2}$, so

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

In (1), each term can be decomposed similarly, and $\frac{1}{k} (k=3, \ldots, n)$ appears three times: twice with a coefficient of $\frac{1}{2}$ and once with a coefficient of -1, the sum of which is clearly 0. $\frac{1}{1}$ and $\frac{1}{n+2}$ appear only once, in the first and last term decompositions, respectively, while $\frac{1}{2}$ and $\frac{1}{n+1}$ appear in the first two and the last two term decompositions, respectively. Thus,

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

Since $\lim _{n \rightarrow \infty} \frac{1}{2 n(n+1)(n+2)}=0$, the sequence is convergent and its limit is $\frac{1}{4}$.

Pethö Attila (Jászberény, Lehel Vezér Gymn., III. grade)

Remark. The decomposition

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

leads to a similar result and even allows for some generalization.

When seeking the limit of the sequence $A_{n}=\frac{1}{1 \cdot 2 \ldots k}+\frac{1}{2 \cdot 3 \ldots(k+1)}+\ldots+\frac{1}{n(n+1) \ldots(n+k-1)}$, we can apply the easily verifiable identity

$$
\begin{gathered}
\frac{1}{n(n+1) \ldots(n+k-1)}=\frac{1}{k-1}\left(\frac{1}{n(n+1) \ldots(n+k-2)}\right)- \\
-\left(\frac{1}{(n+1)(n+2) \ldots(n+k-1)}\right)
\end{gathered}
$$

Using this, we get

$$
A_{n}=\frac{1}{k-1}\left(\frac{1}{(k-1)!}-\frac{1}{(n+1) \ldots(n+k-1)}\right)
$$

Finally, $\lim _{n \rightarrow \infty} A_{n}=\frac{1}{(k-1)(k-1)!}$

Thus,

$$
\sum_{n=1}^{\infty} \frac{n!}{(n+k-1)!}=\frac{1}{(k-1)(k-1)!}
$$

Lenkei Péter (Budapest, József A. Gymn., IV. grade)