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Question Number 142150 by iloveisrael last updated on 27/May/21

\underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{4}}) = ?

Commented by liberty last updated on 27/May/21

\underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{4}}) = - \frac{\sin (\sqrt[4]{- 1} π) \sin (\sqrt[4]{{(- 1)}^{3}} π)}{π^{2}}

Answered by Dwaipayan Shikari last updated on 27/May/21

\underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{4}}) = \underset{n = 1}{\prod^{\infty}} (1 + {(\frac{\sqrt{i}}{n})}^{2}) (1 - {(\frac{\sqrt{i}}{n})}^{2})

= \frac{s i n h (\sqrt{i} π)}{4 π^{2}} . \frac{s i n (\sqrt{i} π)}{i} = - \frac{(e^{\frac{1 + i}{\sqrt{2}} π} - e^{- \frac{1 + i}{\sqrt{2}} π}) (e^{\frac{i}{\sqrt{2}} - \frac{1}{\sqrt{2}} π} - e^{- \frac{i}{\sqrt{2}} + \frac{1}{\sqrt{2}} π})}{4 π^{2}}

= - \frac{e^{\sqrt{2} i π} - e^{- \sqrt{2} π} - e^{\sqrt{2 π}} + e^{- \sqrt{2} i π}}{4 π^{2}} = \frac{c o s h (\sqrt{2} π) - c o s (\sqrt{2} π)}{2 π^{2}}