# How prove that $\sum_{j=1}^{\infty} \prod_{k=1}^j \frac{k-1.5}{k} = -1$ [closed]

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 halvorsenJun 8 '15 at 19:59

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