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Question Number 137307 by mnjuly1970 last updated on 31/Mar/21

$$.....advancedcalculus.....$$$$provethat::$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{{(-1)}^n}{4n^2+1}\right)=\frac{1}{4}(2+\pi csch\left(\frac{\pi }{2}\right))$$$$............................$$

Answered by Dwaipayan Shikari last updated on 31/Mar/21

$$1-\frac{1}{5}+\frac{1}{17}-\frac{1}{37}+\frac{1}{65}-\frac{1}{101}+...$$$$1,17,65$$$$g_n={an}^2+bn+c\Rightarrow g_0=c=1$$$$g_1=a+b+c=17\Rightarrow a+b=16$$$$g_2=4a+2b+1=65\Rightarrow 2a+b=32\Rightarrow a=16b=0c=1\Rightarrow g_n=16n^2+1$$$$5,37,101g_n={(4n+2)}^2+1$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{16n^2+1}-\frac{1}{16n^2+16n+5}$$$$=\frac{1}{16}(\frac{\pi }{2.\frac{1}{4}}coth\left(\frac{\pi }{4}\right)-\frac{1}{2.\frac{1}{16}})-\frac{1}{4i\sqrt{5}}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-\left(\frac{-4+i\sqrt{5}}{8}\right))}-\frac{1}{(n-\left(\frac{-4-i\sqrt{5}}{8}\right))}$$$$=\frac{\pi }{8}coth\left(\frac{\pi }{4}\right)-\frac{1}{2}-\frac{1}{4i\sqrt{5}}(\psi (\frac{1}{2}+\frac{i\sqrt{5}}{4})-\psi (\frac{1}{2}-\frac{i\sqrt{5}}{8}))$$$$=\frac{\pi }{8}coth\left(\frac{\pi }{4}\right)-\frac{1}{2}+\frac{\pi }{4i\sqrt{5}}cot(\frac{1}{2}+\frac{i\sqrt{5}}{8})\pi$$$$=\frac{\pi }{8}coth\left(\frac{\pi }{4}\right)-\frac{1}{2}-\frac{\pi }{4\sqrt{5}}tanh\left(\frac{\sqrt{5}}{8}\pi \right)$$$$:($$

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