Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

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

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


share|cite|improve this question
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. is back. Jun 9 '12 at 18:28

2 Answers 2

up vote 4 down vote accepted

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

share|cite|improve this answer
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 $\, a_0$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.