On multiple occasions, people I know have given me their own lucid expositions of some topic within mathematics, but lack the motivation to write their ideas up properly anywhere. As an expression of my gratitude, and simply because I want to share their explanations with others, I’ve decided to start writing up these “lost stories”. Stay tuned for the first of these, which was motivated by the following problem, posed to me yesterday by Kamal Rai: prove that

$\displaystyle \sum_{k=1}^\infty \frac{1}{k(k+1) \cdots (k+n)} = \frac{1}{n \cdot (n!)}.$

just another guy trying to make the diagrams commute.
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### One Response to

1. shashwat1e4 says:

How does simple induction work on such a problem? I’ve written out an incomplete, half-proof below:

If $n = 1$, then we know that $\sum_{k=1} ^\infty \frac{1}{k(k+1)} = \lim_{p \to \infty} \frac{p}{p+1} = 1$.

The problem, however, arises in the inductive step, where, after some simplification, the proof to the equation $\frac{p^3 - 1}{p(p^2 - 1)(p + 1)!} = \sum_{k=1} ^\infty \frac{(n - 1)! - (n - 2)!}{(p + n)!}$.

The only problem I find is the recursive inductive step; and if this method does not work, why doesn’t it?

Shashwat