Let's calculate the sum of the following series:

$$
\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots+\frac{n}{2^{n-1}}
$$

What does the sum approach as the number of terms grows beyond any bound?
Bring the terms to a common denominator:

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

For now, let's focus on the numerator. We break down the individual terms as follows: $2 \cdot 2^{n-2}=2^{n-2}+2^{n-2}, 3 \cdot 2^{n-3}=$ $2^{n-3}+2^{n-3}+2^{n-3}$, etc., and rearrange the expression:

$$
\begin{gathered}
\left(2^{n-1}+2^{n-2}+\ldots+1\right)+\left(2^{n-2}+2^{n-3}+\ldots+1\right)+ \\
\left(2^{n-3}+\ldots 1\right)+\ldots+(2+1)+1
\end{gathered}
$$

We sum the individual geometric series, and then sum the resulting geometric series:

$$
\begin{gathered}
\frac{2^{n}-1}{2-1}+\frac{2^{n-1}-1}{2-1}+\frac{2^{n-2}-1}{2-1}+\cdots+\frac{2^{2}-1}{2-1}+\frac{2-1}{2-1}= \\
=\frac{2^{n}+2^{n-1}+2^{n-2}+\ldots+2^{2}+2+1-(n+1)}{2-1}= \\
=\frac{2^{n+1}-1}{2-1}-(n+1)=2^{n-1}-(n+2)
\end{gathered}
$$

From this, we get the original expression by dividing by $2^{n-1}$:

$$
\frac{2^{n+1}-(n+2)}{2^{n-1}}=4-\frac{n+2}{2^{n-1}}
$$

In the denominator of the second term, $n$ appears as an exponent, while in the numerator, it appears only as a term. Therefore, the denominator grows much faster than the numerator as $n$ increases, meaning the value of the fraction decreases. If $n$ grows beyond all bounds, this term approaches 0. However, $2^{2} n$ is independent of the value of $n$:

$$
\lim _{n \rightarrow+\infty}\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots+\frac{n}{2^{n-1}}\right)=4
$$

Durst Endre (Szolnok IV. o.)