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What is the exact value of $\prod_{n=1}^{\infty} \left ( 1+\frac{1+i}{n^2+n}\right )$. I wasn't able to find a value on wolfram alpha, but I'm pretty sure a limit exists, though I'm not able to prove it.

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  • $\begingroup$ Have you tried simplifying this into smaller products and evaluating those limits ? Or tried anything at all ? $\endgroup$ – User2648648 Jan 8 '18 at 3:58
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Well, you can obtain that through Gamma function, but in this special case, it's a product of Gamma values, and the well-known functional equation will leave you with a sine. So we can start with that one, as well: $$\frac{\sin\pi z}{\pi z}=\prod^\infty_{n=1}\left(1-\frac{z^2}{n^2}\right)=\prod^\infty_{n=1}\left(1-\frac{z}{n}\right)\left(1+\frac{z}{n}\right)$$ would be the Weierstrass representation of the entire function on the LHS, showing the zeros nicely (the exponential terms cancel out in this case, conveniently). We can multiply that by the telescoping product $$\frac1{1-z}=\prod^\infty_{n=1}\frac{1-\frac{z}{n+1}}{1-\frac{z}{n}}$$ to arrive at $$\frac{\sin\pi z}{\pi z(1-z)}=\prod^\infty_{n=1}\left(1-\frac{z}{n+1}\right)\left(1+\frac{z}{n}\right)=\prod^\infty_{n=1}\left(1+\frac{z(1-z)}{n^2+n}\right).$$ As luck has it, this is exactly what we're looking for, with $z=i$, so we arrive at the same result $\displaystyle\frac{\sinh\pi}{2\pi}(1+i)$, using the well-known formula $\sin ix=i\sinh x$.

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You wrote I wasn't able to find a value on wolfram alpha but this is not my experience.

Assuming that $i=\sqrt{-1}$ and giving Wolfram Alpha

 product (1+(1+i)/(n+n^2)) from n=1 to infinity

I received the partial product $$\prod_{n=1}^{m} \left ( 1+\frac{1+i}{n^2+n}\right )=\frac{\left(\frac{1}{2}+\frac{i}{2}\right) \sinh (\pi )\, \Gamma (m+1+i) \,\Gamma (m+2-i)}{\pi\, \Gamma (m+1)\, \Gamma (m+2)}$$ and making $m\to \infty$, this reduces to $$\prod_{n=1}^{\infty} \left ( 1+\frac{1+i}{n^2+n}\right )=\frac{1+i}{2\pi} \sinh(\pi)\approx 1.83804 +1.83804\, i$$ also given by Wolfram Alpha.

If you consider more generally $$\prod_{n=1}^{\infty} \left ( 1+\frac{a}{n^2+n}\right )=\frac{\cos \left(\frac{\pi}{2} \sqrt{1-4 a}\right)}{\pi a}$$ making $a=1+i$, after simplifications, you get the same result.

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  • $\begingroup$ I see I missed the limits on wolfram alpha, but how does one obtain the partial product, and how is the general formula derived? $\endgroup$ – Abhishek Ab Jan 8 '18 at 7:19
  • $\begingroup$ @AbhishekAb. Google for Infinite products $\endgroup$ – Claude Leibovici Jan 8 '18 at 7:23
  • $\begingroup$ You can get a lot out of Wolfram Alpha, if you know how, very good! It's possible without, too, but just to be on the safe side, I've checked my answer numerically as well (with GP, because I was offline, on the bus). $\endgroup$ – Professor Vector Jan 8 '18 at 7:59

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