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1-2-pi-2-1-2-pi-2-1-6-pi-2-1-6-pi-2-1-10-pi-2-1-10-pi-2-pi-2-16-sec-2-pi-2-4-Prove-or-disprove-

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Question Number 130555 by Dwaipayan Shikari last updated on 26/Jan/21
(1/((2−π)^2 ))+(1/((2+π)^2 ))+(1/((6−π)^2 ))+(1/((6+π)^2 ))+(1/((10−π)^2 ))+(1/((10+π)^2 ))+..=(π^2 /(16))sec^2 ((π^2 /4)) Prove or disprove
$$\frac{\mathrm{1}}{\left(\mathrm{2}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}+\pi\right)^{\mathrm{2}} }+..=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}{sec}^{\mathrm{2}} \left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${Prove}\:{or}\:{disprove} \\ $$
Answered by mindispower last updated on 27/Jan/21
S=Σ_(k≥0) (1/((4k+2−π)^2 ))+(1/((4k+2+π)^2 ))=(1/(16))Σ_(k≥0) (1/((k+((2−π)/4))^2 ))+(1/((k+((2+π)/4))^2 )) cot(x)=(1/x)+Σ_(n≥1) (1/((x−πn)))+(1/((x+πn))) ⇒tg(πx)=cot((π/2)−x)=(2/(π−2x))+Σ_(n≥1) (2/(π−2x−2πn))+(2/(π−2x+2πn)) ⇒1+tg^2 (x)=(4/((π−2x)^2 ))+Σ_(n≥1) (4/((π−2x−2πn)^2 ))+(4/((π−2x+2πn)^2 ))=H x=(π^2 /4) ⇒H=(4/((π−(π^2 /2))^2 ))+Σ_(n≥1) (4/((π−(π^2 /2)−2πn)))+(4/((π−(π^2 /2)+2πn)^2 ))=A+B B=Σ_(n≥0) {(4/((−π−(π^2 /2)−2nπ)^2 ))+(4/((π+2πn−(π^2 /2))^2 ))}−(4/((π−(π^2 /2))^2 )) =(1/π^2 )Σ_(n≥0) (1/((((2+π)/4)+n)^2 ))+(1/((n+((2−π)/4))^2 ))−(4/((π−(π^2 /2)))) =((16)/π^2 )S−(4/((π−(π^2 /2))^2 ))=sec^2 ((π^2 /4))−(4/((π−(π^2 /2))^2 )) ⇒S=(π^2 /(16))sec^2 ((π^2 /4))
$${S}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{4}{k}+\mathrm{2}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{4}{k}+\mathrm{2}+\pi\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{2}−\pi}{\mathrm{4}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{2}+\pi}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$${cot}\left({x}\right)=\frac{\mathrm{1}}{{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({x}−\pi{n}\right)}+\frac{\mathrm{1}}{\left({x}+\pi{n}\right)} \\ $$$$\Rightarrow{tg}\left(\pi{x}\right)={cot}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{2}}{\pi−\mathrm{2}{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}}{\pi−\mathrm{2}{x}−\mathrm{2}\pi{n}}+\frac{\mathrm{2}}{\pi−\mathrm{2}{x}+\mathrm{2}\pi{n}} \\ $$$$\Rightarrow\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{4}}{\left(\pi−\mathrm{2}{x}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{4}}{\left(\pi−\mathrm{2}{x}−\mathrm{2}\pi{n}\right)^{\mathrm{2}} }+\frac{\mathrm{4}}{\left(\pi−\mathrm{2}{x}+\mathrm{2}\pi{n}\right)^{\mathrm{2}} }={H} \\ $$$${x}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{H}=\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\pi{n}\right)}+\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\pi{n}\right)^{\mathrm{2}} }={A}+{B} \\ $$$${B}=\underset{{n}\geqslant\mathrm{0}} {\sum}\left\{\frac{\mathrm{4}}{\left(−\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{n}\pi\right)^{\mathrm{2}} }+\frac{\mathrm{4}}{\left(\pi+\mathrm{2}\pi{n}−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }\right\}−\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\frac{\mathrm{2}+\pi}{\mathrm{4}}+{n}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{2}−\pi}{\mathrm{4}}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{16}}{\pi^{\mathrm{2}} }{S}−\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }={sec}^{\mathrm{2}} \left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)−\frac{\mathrm{4}}{\left(\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{S}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}{sec}^{\mathrm{2}} \left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 27/Jan/21
Great sir! I had posted this problem on Brilliant.You could check their also
$${Great}\:{sir}!\:{I}\:{had}\:{posted}\:{this}\:{problem}\:{on}\:{Brilliant}.{You}\:{could} \\ $$$${check}\:{their}\:{also} \\ $$
Commented by mindispower last updated on 27/Jan/21
withe pleasur sir i have not yet Brilliant
$${withe}\:{pleasur}\:{sir}\:{i}\:{have}\:{not}\:{yet}\:{Brilliant} \\ $$$$ \\ $$

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