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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
1(2−π)2+1(2+π)2+1(6−π)2+1(6+π)2+1(10−π)2+1(10+π)2+..=π216sec2(π24)Proveordisprove
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=∑k⩾01(4k+2−π)2+1(4k+2+π)2=116∑k⩾01(k+2−π4)2+1(k+2+π4)2cot(x)=1x+∑n⩾11(x−πn)+1(x+πn)⇒tg(πx)=cot(π2−x)=2π−2x+∑n⩾12π−2x−2πn+2π−2x+2πn⇒1+tg2(x)=4(π−2x)2+∑n⩾14(π−2x−2πn)2+4(π−2x+2πn)2=Hx=π24⇒H=4(π−π22)2+∑n⩾14(π−π22−2πn)+4(π−π22+2πn)2=A+BB=∑n⩾0{4(−π−π22−2nπ)2+4(π+2πn−π22)2}−4(π−π22)2=1π2∑n⩾01(2+π4+n)2+1(n+2−π4)2−4(π−π22)=16π2S−4(π−π22)2=sec2(π24)−4(π−π22)2⇒S=π216sec2(π24)
Commented by Dwaipayan Shikari last updated on 27/Jan/21
Great sir! I had posted this problem on Brilliant.You could check their also
Greatsir!IhadpostedthisproblemonBrilliant.Youcouldchecktheiralso
Commented by mindispower last updated on 27/Jan/21
withe pleasur sir i have not yet Brilliant
withepleasursirihavenotyetBrilliant

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