Kvant Math Problem 809

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Problem

Find the sum $$\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{n-1}{n!}$$ (the product $1\cdot2\cdot\ldots\cdot k$ is denoted by $k!$).

V. V. Proizvolov

Exploration

Compute the first few terms of the series to detect a pattern. For $n=2$, the sum is $\frac{1}{2!} = \frac{1}{2}$. For $n=3$, the sum is $\frac{1}{2!} + \frac{2}{3!} = \frac{1}{2} + \frac{2}{6} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. For $n=4$, the sum is $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} = \frac{12}{24} + \frac{8}{24} + \frac{3}{24} = \frac{23}{24}$. For $n=5$, the sum is $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{30} = \frac{23}{24} + \frac{1}{30} = \frac{115/120 + 4/120} = \frac{119}{120}$.

The sums approach 1 as $n$ increases. Each term can be rewritten as $\frac{k}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$, suggesting a telescoping structure. The crucial point is verifying that this decomposition is correct and that the telescoping works rigorously for all $k$.

Problem Understanding

We are asked to find the sum

$S_n = \frac{1}{2!} + \frac{2}{3!} + \cdots + \frac{n-1}{n!}.$

This is a Type A problem because it asks to find the sum in closed form. The core difficulty is identifying the correct manipulation of each term to reveal a telescoping pattern. The preliminary exploration suggests that

$\frac{k}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!},$

which produces a telescoping sum that collapses neatly to a simple expression. Intuitively, the sum should be

$S_n = 1 - \frac{1}{n!},$

because the pattern of differences leaves only the first and last terms.

Proof Architecture

Lemma 1: For every integer $k \ge 2$, the equality $\frac{k}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$ holds. This is true because

$\frac{1}{(k-1)!} - \frac{1}{k!} = \frac{k}{k \cdot (k-1)!} - \frac{1}{k!} = \frac{k-1}{k!}$

and we need to verify carefully. The hardest step is confirming the decomposition without error.

Lemma 2: The sum $\sum_{k=2}^n \left(\frac{1}{(k-1)!} - \frac{1}{k!}\right)$ telescopes to $1 - \frac{1}{n!}$. This is true because all intermediate terms cancel in a consecutive difference sequence.

Main Claim: The sum $S_n$ equals $1 - \frac{1}{n!}$, by combining Lemmas 1 and 2. The main risk is failing to handle the first or last term correctly in the telescoping.

Solution

We first prove Lemma 1. For any integer $k \ge 2$, we compute:

$\frac{1}{(k-1)!} - \frac{1}{k!} = \frac{k}{k \cdot (k-1)!} - \frac{1}{k!} = \frac{k - 1}{k!}.$

Multiplying and simplifying carefully,

$\frac{k}{k!} = \frac{k}{k \cdot (k-1)!} = \frac{1}{(k-1)!},$

$\frac{1}{(k-1)!} - \frac{1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k \cdot (k-1)!} = \frac{k-1}{k \cdot (k-1)!} = \frac{k-1}{k!}.$

This confirms the decomposition is correct for $k \ge 2$.

Applying Lemma 1 to each term of $S_n$, we obtain

$S_n = \sum_{k=2}^n \frac{k-1}{k!} = \sum_{k=2}^n \left(\frac{1}{(k-1)!} - \frac{1}{k!}\right).$

Reindexing the first part of the sum, let $j = k-1$, then $\sum_{k=2}^n \frac{1}{(k-1)!} = \sum_{j=1}^{n-1} \frac{1}{j!}$. Therefore,

$S_n = \sum_{j=1}^{n-1} \frac{1}{j!} - \sum_{k=2}^n \frac{1}{k!}.$

Combining the two sums, we note that all terms cancel except the first term of the first sum, $\frac{1}{1!} = 1$, and the last term of the second sum, $-\frac{1}{n!}$. Hence,

$S_n = 1 - \frac{1}{n!}.$

This completes the proof. ∎

Verification of Key Steps

Re-examine Lemma 1 carefully. For $k=2$, $\frac{2}{2!} = 1$ and $\frac{1}{1!} - \frac{1}{2!} = 1 - \frac{1}{2} = \frac{1}{2}$. There is a miscalculation. Recompute:

$\frac{1}{(k-1)!} - \frac{1}{k!} = \frac{1}{1!} - \frac{1}{2!} = 1 - \frac{1}{2} = \frac{1}{2},$

which matches $\frac{k-1}{k!} = \frac{1}{2!} = \frac{1}{2}$. Therefore the formula should be $\frac{k-1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$. This is the correct decomposition.

Recheck the telescoping sum with $n=4$: $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} = \frac{23}{24}$. Using formula $1 - \frac{1}{4!} = 1 - \frac{1}{24} = \frac{23}{24}$. Verified.

Alternative Approaches

An alternative approach is to note that $\frac{k-1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$ and directly write $S_n = \sum_{k=2}^n \frac{1}{(k-1)!} - \sum_{k=2}^n \frac{1}{k!} = \sum_{j=1}^{n-1} \frac{1}{j!} - \sum_{k=2}^n \frac{1}{k!} = 1 - \frac{1}{n!}$ without introducing the telescoping terminology. The main approach is preferable because it clearly explains why the intermediate terms vanish and highlights the telescoping structure, making the underlying pattern transparent.