Let's determine the sum of the following series:

$$
1+2 \cdot x+3 \cdot x^{2}+\ldots+n \cdot x^{n-1}+(n+1) \cdot x^{n}
$$

What is the sum when $x<1$ and $n$ is infinitely large?
The given series is the sum of the following geometric progressions:

$$
\begin{array}{rccc}
1+x+x^{2}+x^{3}+ & \ldots & +x^{n} \\
x+x^{2}+x^{3}+ & \cdots & +x^{n} \\
x^{2}+x^{3}+ & \cdots & +x^{n} \\
\cdots & \cdots & \cdots & \cdots \\
& \cdots & \cdots & \cdots \\
& & x^{n-1}+ & x^{n} \\
& & & x^{n}
\end{array}
$$

Thus,

$$
S=\frac{x^{n+1}-1}{x-1}+\frac{x^{n+1}-x}{x-1}+\frac{x^{n+1}-x^{2}}{x-1}+\ldots+\frac{x^{n+1}-x^{n}}{x-1}
$$

Or

$$
S=\frac{(n+1) x^{n+1}-\left(1+x+x^{2}+\ldots+x^{n}\right)}{x-1}=\frac{(n+1) x^{n+2}-(n+2) x^{n+1}+1}{(x-1)^{2}}
$$

If $x<1$ and $n=\infty$, then in (1) the individual terms are:

$$
\frac{1}{1-x}, \quad \frac{x}{1-x}, \quad \frac{x^{2}}{1-x}, \cdots \frac{x^{n}}{1-x}
$$

Thus,

$$
S=\frac{1+x+x^{2}+\ldots}{1-x}=\frac{1}{(1-x)^{2}}
$$

(Viktor Wáhl, Eger.)

The problem was also solved by: Bánó L., Csada I., Ehrenfeld N., Epstein K., Fekete M., Fodor H., Haar A., Heimlich P., Kirchknopf E., Kiss J., Merse P., Paunz A., Pichler S., Ruvald S., Schuster Gy., Schwarz O., Székely J., Tandlich E., Tóth B., Tóth J.