How prove that

$$\sum_{j=1}^{\infty} \prod_{k=1}^j \frac{k-1.5}{k} = -1$$

I have any idea, so any help wil be helpfull.


closed as off-topic by Did, Claude Leibovici, TravisJ, graydad, kjetil b halvorsen Jun 8 '15 at 19:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Claude Leibovici, TravisJ, graydad, kjetil b halvorsen
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ In full generality, $$\sum_{j=1}^\infty\prod_{k=1}^j\frac{k-c}k=-1+\prod_{j=1}^\infty\left(1-\frac{c-1}j\right)$$ hence, for every $c>1$, $$\sum_{j=1}^\infty\prod_{k=1}^j\frac{k-c}k=-1$$ No gamma or beta function or irrational number is involved in the proof, which is purely algebraic, since, for every positive $n$, $$\sum_{j=1}^n\prod_{k=1}^j\frac{k-c}k=-1+\prod_{j=1}^n\left(1-\frac{c-1}j\right)$$ $\endgroup$ – Did Jan 16 '17 at 16:33

Since: $$\prod_{k=1}^{j}\frac{k-\frac{3}{2}}{k}=-\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(j-\frac{1}{2}\right)}{\pi\,\Gamma\left(j+1\right)}=-\frac{1}{\pi}\,B\left(j-\frac{1}{2},\frac{3}{2}\right)\tag{1}$$ we have: $$\begin{eqnarray*}\sum_{j\geq 1}\prod_{k=1}^{j}\frac{k-\frac{3}{2}}{k}=-\frac{1}{\pi}\sum_{j\geq 1}\int_{0}^{1}x^{j-\frac{3}{2}}(1-x)^{\frac{1}{2}}\,dx=-\frac{1}{\pi}\int_{0}^{1}x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\,dx \tag{2}\end{eqnarray*}$$ and: $$\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}}=\Gamma\left(\frac{1}{2}\right)^2=\pi,\tag{3}$$ proving your claim.

  • $\begingroup$ Why $-\frac{1}{\pi}\sum_{j\geq 1}\int_{0}^{1}x^{j-\frac{3}{2}}(1-x)^{\frac{1}{2}}\,dx=-\frac{1}{\pi}\int_{0}^{1}x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\,dx $ $\endgroup$ – Thomas Jun 8 '15 at 16:54
  • $\begingroup$ @Thomas: I simply took the sum of a geometric series, namely $\sum_{j\geq 1}x^{j-1}$. $\endgroup$ – Jack D'Aurizio Jun 8 '15 at 16:57

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