I know that this sum $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)\cdots(n+p)}$$ ($p$ fixed) converges which can be easily proved using the ratio criterion, but I couldn't calculate it.

I need help in this part.

Thanks a lot.

  • $\begingroup$ for $p=0$ the series won't converge, for every other case i would try to part the fractions $\endgroup$ – Dominic Michaelis Feb 18 '13 at 8:16
  • 2
    $\begingroup$ For $p=1$ the series telescopes: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Try to see if this idea extends? $\endgroup$ – Henno Brandsma Feb 18 '13 at 8:17
  • $\begingroup$ yeah it works, $$\frac{1}{p^2\cdot (n-1)!}$$ $\endgroup$ – Dominic Michaelis Feb 18 '13 at 8:19
  • $\begingroup$ This is of the form $x^{(p)}$ which has a well known sum. $\endgroup$ – Ishan Banerjee Feb 18 '13 at 8:25
  • $\begingroup$ @IshanBanerjee What do you mean? which series is of this form? $\endgroup$ – user42912 Feb 18 '13 at 8:29

Hint: $$\frac{p}{n(n+1)\cdots(n+p)}=\frac{1}{(n)(n+1)\cdots (n+p-1)}-\frac{1}{(n+1)(n+2)\cdots (n+p)}.$$

  • $\begingroup$ ah, beat me to it. $\endgroup$ – cats Feb 18 '13 at 8:21

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