1
$\begingroup$

First let's take a look at this identity: $$\displaystyle \sum_{k=1}^{\infty} \frac{1}{2k(2k-1)} = \ln{2} $$ It can be proved easily.

Now consider this generalized infinite series ($m$ is a positive integer greater than $1$ and its value is specified): $$\displaystyle \sum_{k=1}^{\infty} \frac{1}{\displaystyle \prod_{i=1}^{m} (mk-m+i)} $$

This problem may be difficult. Any help will be appreciated.

  • 2
    Partial fraction expansion2017-01-25
  • 0
    That, and roots of unity.2017-01-25
  • 0
    This is a very beautiful question, which basically asks us to find a general formula for the infinite series $$S_m~=~\dfrac1{m!}~\sum_{k=1}^\infty{mk\choose m}^{-1}$$ Some values of $m!~S_m$ can be found [here](http://i.stack.imgur.com/dS5TX.png).2017-01-25

0 Answers 0