# How do we calculate this sum $\sum_{n=1}^{\infty} \frac{1}{n(n+1)\cdots(n+p)}$?

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.

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

## 1 Answer

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

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