Problem 1458

To prove the statement, we will use mathematical induction.

1. Base Case:
For $n = 1$, the only non-empty subset of $\{1\}$ is $\{1\}$ itself. Thus, the sum is:
$$\sum \frac{1}{i_1} = \frac{1}{1} = 1$$
Therefore, the statement holds for $n = 1$.

2. Inductive Step:
Assume the statement is true for $n = k$. That is, we assume:
$$\sum_{\substack{\{i_1, i_2, \ldots, i_j\} \subseteq \{1, 2, \ldots, k\} \\ \{i_1, i_2, \ldots, i_j\} \neq \emptyset}} \frac{1}{i_1 i_2 \ldots i_j} = k$$
We need to show that the statement holds for $n = k+1$. Consider the set $\{1, 2, \ldots, k+1\}$.

3. **Sum over subsets of $\{1, 2, \ldots, k+1\}$:**
The non-empty subsets of $\{1, 2, \ldots, k+1\}$ can be divided into two categories:
- Subsets that do not contain $k+1$
- Subsets that contain $k+1$

4. **Subsets that do not contain $k+1$:**
These are exactly the non-empty subsets of $\{1, 2, \ldots, k\}$. By the induction hypothesis, the sum over these subsets is $k$.

5. **Subsets that contain $k+1$:**
For a subset $S$ that contains $k+1$, we can write $S = T \cup \{k+1\}$, where $T$ is a subset of $\{1, 2, \ldots, k\}$. The sum over these subsets is:
$$\sum_{\substack{T \subseteq \{1, 2, \ldots, k\} \\ T \neq \emptyset}} \frac{1}{(i_1 i_2 \ldots i_j)(k+1)} = \frac{1}{k+1} \sum_{\substack{T \subseteq \{1, 2, \ldots, k\} \\ T \neq \emptyset}} \frac{1}{i_1 i_2 \ldots i_j}$$
By the induction hypothesis, the sum over all non-empty subsets $T$ of $\{1, 2, \ldots, k\}$ is $k$. Therefore, the sum over subsets containing $k+1$ is:
$$\frac{1}{k+1} \cdot k = \frac{k}{k+1}$$

6. Combining the sums:
The total sum over all non-empty subsets of $\{1, 2, \ldots, k+1\}$ is the sum of the two parts:
$$k + \frac{k}{k+1}$$
Simplifying this expression:
$$k + \frac{k}{k+1} = \frac{k(k+1)}{k+1} + \frac{k}{k+1} = \frac{k(k+1) + k}{k+1} = \frac{k^2 + k + k}{k+1} = \frac{k^2 + 2k}{k+1} = \frac{k(k+2)}{k+1} = k + 1$$

Thus, the statement holds for $n = k+1$.

By the principle of mathematical induction, the statement is true for all $n \geq 1$.

$\blacksquare$