Is it true that

$$
\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots+\frac{30}{2^{30}}<2 ?
$$
Is it true that

$$
\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots+\frac{30}{2^{30}}<2 ?
$$

Solution. Let's denote the sum of the first $n$ terms on the left side of (1) by $S_{n}$, and write down a few initial values in the following form:

$$
\begin{aligned}
& S_{1}=\frac{1}{2}=2-\frac{1+2}{2} \\
& S_{2}=\frac{1}{2}+\frac{2}{2^{2}}=2-\frac{2+2}{2^{2}} \\
& S_{3}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}=2-\frac{3+2}{2^{3}}
\end{aligned}
$$

We conjecture that this is generally true, that is, $S_{n}=2-\frac{n+2}{2^{n}}$. We will prove this by mathematical induction.

For $n=1$, our statement is true. We need to show that the property is inherited, that is, if it is true for $k$, then it is also true for $(k+1)$.

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

This proves that our statement is valid for all natural numbers $n$; specifically, $S_{30}=2-\frac{32}{2^{31}}$, which is indeed less than 2.

Based on the work of Kornél Takács (Győr, Pattantyús Á. G. Industrial School II. grade)

Remarks. 1. The problem can be solved in several ways. Some of the solvers broke down the sum into 30 geometric series and summed them up.

2. Some converted the fractions into binary decimals and performed the summation this way.
3. It is also easy to prove a more general form of the inequality. Let $n$ and $k$ be natural numbers; then

$$
\frac{k}{2}+\frac{k+1}{2^{2}}+\ldots+\frac{k+n}{2^{n+1}}<k+1
$$

Multiply the inequality by 2 and subtract the integer $k$ on the left side; then on the right side we get $2(k+1)-k=(k+2)$. Then multiply by 2 again and subtract $(k+1)$; the right side becomes $k+3$. Continuing this process, in the $(n+1)$-th step we arrive at the inequality $0<k+n+2$, which is obviously true. Since we performed equivalent transformations throughout, our original inequality is also true.