To be calculated

$$
S_{n}=\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n+1)(2 n+3)}
$$

What is $\lim S_{n}$, if $n \rightarrow \infty$?
I. Solution. Let's decompose the terms of the series $S_{n}$:

$$
\begin{gathered}
\frac{1}{(2 n+1)(2 n+3)}=\frac{A}{2 n+1}+\frac{B}{2 n+3} \\
1=A(2 n+3)+B(2 n+1)=2(A+B) n+3 A+B
\end{gathered}
$$

This equation must hold for all integer values of $n$; this is only possible if

$$
A+B=0 \text { and } 3 A+B=1, \text { i.e., } A=\frac{1}{2} \text { and } B=-\frac{1}{2}
$$

Thus,

$$
\frac{1}{(2 n+1)(2 n+3)}=\frac{1}{2}\left[\frac{1}{2 n+1}-\frac{1}{2 n+3}\right]=\frac{1}{2}[f(n)-f(n-1)]
$$

if, i.e.,

$$
f(n)=\frac{1}{2 n+1}
$$

Accordingly,

$$
\begin{gathered}
\frac{1}{1 \cdot 3}=\frac{1}{2}[f(0)-f(1)] \\
\frac{1}{3 \cdot 5}=\frac{1}{2}[f(1)-f(2)] \\
\frac{1}{5 \cdot 7}=\frac{1}{2}[f(2)-f(3)] \\
\\
\cdots \cdots \\
\frac{1}{(2 n+1)(2 n+3)}=\frac{1}{2}[f(n)-f(n+1)]
\end{gathered}
$$

thus,

$$
S_{n}=\frac{1}{2}[f(0)-f(n+1)]=\frac{1}{2}\left(1-\frac{1}{2 n+3}\right)=\frac{n+1}{2 n+3}
$$

and

$$
\text { if } n \rightarrow \infty, \text { then } \lim S_{n}=\lim \frac{1+\frac{1}{n}}{2+\frac{3}{n}}=\frac{1}{2}
$$

Miklós Tóth (Premontrei g. VII. o. Gödöllő.)

II. Solution. The calculation of the sum $S_{n}$ can also be done based on the fact that

$$
\frac{1}{(2 n+1)(2 n+3)}=\frac{n+1}{2 n+1}-\frac{n+2}{2 n+3}
$$

III. Solution. The calculation of the sum $S_{n}$ was performed by complete induction:

Gy. Bálint, E. Bluszt, Z. Marosán, V. Csuri, G. Donáth, V. Frida, I. Füstös, R. Gáspár, S. Gombos, L. Orosz, P. Holczer, F. Kieweg, Gy. Királyhídi, I. Koch, J. Krisztonosich, G. Lázár, L. Major, B. Mandl, E. Németh, T. Pappert, L. Szentmiklósi, L. Szerényi, R. Szkitsák, J. Szabó, F. Tasnádi, A. Weisz, P. Zubek.