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I am trying to sum the series

$ \sum u_{n}$

where $ u_{n}=\frac{a+n-1}{\prod_{j=1}^n (a+j)}$ $ a>0$ We have:

$ \frac{a+n-1}{\prod_{j=1}^n (a+j)}=\sum_{k=1}^n\frac{b_k}{a+k} $

$ b_{k}=\frac{n-k-1}{\prod_{j=1,j\neq k}^n (j-k)}$

$ \sum_{n=1}^N u_{n}= \sum_{n=1}^N \sum_{k=1}^n\frac{n-k-1}{(a+k)\prod_{j=1,j\neq k}^n (j-k)}$

...

  • 2
    More compactly, your sum is $\sum_{k=1}^\infty \frac{a+k-1}{(a+1)_k}$, where $(a)_k$ is the Pochhammer symbol.2012-06-09

2 Answers 2

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Since $(x)_n = \frac{ \Gamma(x+n)}{\Gamma(x)} $ your sum can be written as $\sum_{k=1}^{\infty} \frac{(a+k-1)\Gamma(a+1)}{\Gamma(a+k+1)} .$

Note $\frac{a+k-1}{\Gamma(a+k+1)} = \frac{a+k}{(a+k)\Gamma(a+k)} - \frac{1}{\Gamma(a+k+1)}= \frac{1}{\Gamma(a+k)} - \frac{1}{\Gamma(a+k+1)}.$

So $\sum_{k=1}^{\infty} \frac{(a+k-1)\Gamma(a+1)}{\Gamma(a+k+1)} =1.$

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    Ok, thank you very much Ragib Zaman!2019-01-25
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You should make it a telescopic sum. like this: $u_n=a_{n-1}-a_n\,$ where $a_n=\frac{1}{ \prod_{j=1}^{n}(a+j)} $ and easily $\lim_{n \rightarrow \infty} a_n =0\,$ hence your sum is $\, a_1 $ or $\, a_0$