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The problem is to prove that :

$\frac{2}{0!+1!+2!} + \frac{3}{1!+2!+3!} + \dots + \frac{n}{(n-2)!+(n-1)!+n!} = 1 - \frac{1}{n!}$

I used the following rather lengthy approach to prove this:

The general term $\frac{k}{(k-2)!+(k-1)!+k!}$ can be simplified to $\frac{k-1}{k!}$.

I induct on the proposition $P(k): \frac{1}{n!} + \sum\limits_{k =1}^{n-1} \frac{k}{(k+1)!} = 1$, for integer $k > 1$. This establishes the result but I am looking for a quicker way to prove this, which will probably require some clever algebraic trickery.

Thanks.

erdoswiles
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1 Answers1

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You're so close already! Rearrange from what you have to get $$\sum_{k=2}^{n} \frac{k-1}{k!} = \sum_{k=2}^{n}\left( \frac{1}{(k-1)!} - \frac{1}{k!}\right)$$ which is a nice telescoping series. So, it immediately drops out that $$\sum_{k=2}^{n} \frac{k-1}{k!} = \frac{1}{(2-1)!} - \frac{1}{n!}$$ as required.

B. Mehta
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