0
$\begingroup$

How to find the sum of this series by partial fraction ,I am not able to apply partial fraction here .$\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)\cdots(n+m)}$.Thanks for help in advance

$\endgroup$
0

2 Answers 2

3
$\begingroup$

Assuming that your running index is $n$ instead of $i$, here is a simple partial fraction trick

$$ \frac{1}{n(n+1)\cdots(n+m)} = \frac{1}{m} \left( \frac{1}{n(n+1)\cdots(n+m-1)} - \frac{1}{(n+1)\cdots(n+m)} \right) $$

from which you can initiate the telescoping argument.

$\endgroup$
2
$\begingroup$

\begin{eqnarray*} &\frac{1}{n(n+1) \cdots (n+m-1)(n+m)} \\ =& \frac{1/m}{n(n+1) \cdots (n+m-1)}-\frac{1/m}{(n+1) \cdots (n+m-1)(n+m)} \end{eqnarray*} & then do a telescoping sum.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .