Unexpected Sum

Problem

Find the exact value of the following infinite series:

$$\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dfrac{4}{5!} + ...$$

Solution

First we note that the general term in this series can be written differently:

$$\dfrac{k}{(k+1)!} = \dfrac{k + 1 - 1}{(k+1)!} = \dfrac{k + 1}{(k+1)!} - \dfrac{1}{(k+1)!} = \dfrac{1}{k!} - \dfrac{1}{(k+1)!}$$

Hence the original series becomes a telescoping series:

\begin{align}\dfrac{1}{2!} + \dfrac{2}{3!} + ... &= \left( \dfrac{1}{1!} - \dfrac{1}{2!} \right) + \left( \dfrac{1}{2!} - \dfrac{1}{3!} \right) + \left( \dfrac{1}{3!} - \dfrac{1}{4!} \right) + ...\\&= 1\end{align}
Problem ID: 275 (21 Apr 2006)     Difficulty: 3 Star

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