# Summing the series $\sum u_{n}, u_{n}=\frac{a+n-1}{(a+1)…(a+n)}$

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:

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

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

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

...

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More compactly, your sum is $$\sum_{k=1}^\infty \frac{a+k-1}{(a+1)_k}$$, where $(a)_k$ is the Pochhammer symbol. – J. M. Jun 9 '12 at 18:28

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! – Chon Jun 9 '12 at 21:22

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 $\, a_0$

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