Some Values .. Σ_(n=−∞) ^∞ e^(−πn^2 ) =(π^(1/4) /(Γ((3/4)))) Σ_(n=−∞) ^∞ e^(−2πn^2 ) =(π^(1/4) /(Γ((3/4)))) (((6+4(√2)))^(1/4) /2) Σ_(n=−∞) ^∞ e^(−6πn^2 ) =(π^(1/4) /(Γ((3/4)))).((√((1)^(1/4) +(3)^(1/4) +(4)^(1/4) +(9)^(1/4) ))/( (√(1728)))) Any Idea to prove ?


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Question Number 127982 by Dwaipayan Shikari last updated on 03/Jan/21

$S o m e V a l u e s . .$ $\underset{n = - \infty}{\sum^{\infty}} e^{- π n^{2}} = \frac{π^{\frac{1}{4}}}{Γ (\frac{3}{4})}$ $\underset{n = - \infty}{\sum^{\infty}} e^{- 2 π n^{2}} = \frac{π^{\frac{1}{4}}}{Γ (\frac{3}{4})} \frac{\sqrt[4]{6 + 4 \sqrt{2}}}{2}$ $\underset{n = - \infty}{\sum^{\infty}} e^{- 6 π n^{2}} = \frac{π^{\frac{1}{4}}}{Γ (\frac{3}{4})} . \frac{\sqrt{\sqrt[4]{1} + \sqrt[4]{3} + \sqrt[4]{4} + \sqrt[4]{9}}}{\sqrt{1728}}$ $A n y I d e a t o p r o v e ?$
Commented by Dwaipayan Shikari last updated on 03/Jan/21

$T h e r e a r e s o m a n y t h e t a f u n c t i o n v a l u e s d e r i v e d b y R a m a n u j a n$ $\underset{n = - \infty}{\sum^{\infty}} e^{- 3 π n^{2}} = \frac{\sqrt[4]{π}}{Γ (\frac{3}{4})} . \frac{\sqrt[4]{27 + 18 \sqrt{3}}}{3}$ $\underset{n = - \infty}{\sum^{\infty}} e^{- 4 π n^{2}} = \frac{\sqrt[4]{π}}{Γ (\frac{3}{4})} . \frac{\sqrt[4]{8} + 2}{4}$ $\underset{n = - \infty}{\sum^{\infty}} e^{- 7 π n^{2}} = \frac{\sqrt[4]{π}}{Γ (\frac{3}{4})} . \frac{\sqrt{7 + 4 \sqrt{7} + 5 \sqrt[4]{28} + \sqrt[4]{1378}}}{\sqrt{7}}$
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