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Question Number 147474 by mnjuly1970 last updated on 21/Jul/21

Answered by mindispower last updated on 21/Jul/21

$$\frac{1}{2a}(\underset{n⩾-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{4n-a}-\frac{{(-1)}^n}{4n+a})=\sum _{n⩾-\mathrm{\infty }}\frac{{(-1)}^n}{16n^2-a^2}=f\left(a\right)$$$$=\frac{1}{2a}\sum _{n⩾-\mathrm{\infty }}\frac{1}{8n-a}-\frac{1}{8n-a+4}-\frac{1}{8n+a}+\frac{1}{8n+a+4}$$$$recall\sum _{n⩾-\mathrm{\infty }}\frac{1}{x+n}=\pi cot\left(\pi x\right)$$$$f\left(a\right)=\frac{\pi }{16a}(cot(-\frac{\pi a}{8})-cot\left(\frac{4-a}{8}\pi \right)-cot\left(\frac{\pi a}{8}\right)+cot\left(\frac{4+a}{8}\pi \right))$$$$=-\frac{\pi }{8a}(cot\left(\frac{\pi a}{8}\right)+tan\left(\frac{\pi a}{8}\right))$$$$cot\left(a\right)+tan\left(a\right)=\frac{2}{sin\left(2a\right)}$$$$=-\frac{\pi }{4asin\left(\frac{\pi a}{4}\right)}$$$$\left(1\right)=f\left(1\right)=\frac{\pi }{4sin\left(\frac{\pi }{4}\right)}=-\frac{\pi }{4}\sqrt{2}$$$$\left(2\right)f\left(1\right)=2S-1\Rightarrow s=\frac{1}{2}-\frac{\pi }{4\sqrt{2}}$$$$\left(3\right)f^{\prime }\left(a\right)=\sum _{n⩾-\mathrm{\infty }}\frac{2a{(-1)}^n}{{(16n^2-a^2)}^2}$$$$S=\frac{f^{\prime }\left(1\right)}{2}$$$$f^{\prime }\left(a\right)=\frac{\pi }{4}\left(\frac{sin\left(\frac{\pi a}{4}\right)+\frac{\pi }{4}acos\left(\frac{\pi }{4}a\right)}{{\left(asin\left(\frac{\pi }{4}a\right)\right)}^2}\right)$$$$f^{\prime }\left(1\right)=\frac{\pi }{4}(\sqrt{2}+\frac{\pi }{4}\sqrt{2})=\frac{{\pi }^2}{16}\sqrt{2}+\frac{\pi \sqrt{2}}{4}$$$$S=\frac{{\pi }^2}{32}\sqrt{2}+\frac{\pi \sqrt{2}}{8}$$$$\left(4\right),S=2S^{\prime }+1$$$$S^{\prime }=\frac{{\pi }^2}{64}\sqrt{2}+\frac{\pi \sqrt{2}}{16}-\frac{1}{2}$$$$$$$$$$$$$$$$$$$$$$

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