Intereting Posts

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Is there a general formula for

$\sum_{k=1}^n\frac{1}{k(k+1)…(k+m)}$?

I know that the limit is $\frac{1}{mm!}$ but is there a combinatorial expression for this?

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$$

\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m)} = \frac1m\sum_{k=0}^n \frac{(k + m) – k}{k(k+1)\cdots(k+m)} \\

= \frac1m\sum_{k=1}^n \left[ \frac{1}{k(k+1)\cdots(k+m-1)} – \frac{1}{(k+1)(k+2)\cdots(k+m)}\right]

$$

This is a telescoping sum. The result then is

$$

-\frac1m\left[ \frac{1}{(n+1)(n+2)\cdots(n+m)} – \frac{1}{1\cdot 2\cdots m}\right ] \\

= -\frac1m\left[ \frac{n!}{(m+n)!} – \frac{1}{m!}\right] \\

= \frac{1}{mm!} – \frac{n!}{m(m+n)!}

$$

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