(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


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Question Number 130555 by Dwaipayan Shikari last updated on 26/Jan/21

$\frac{1}{{(2 - π)}^{2}} + \frac{1}{{(2 + π)}^{2}} + \frac{1}{{(6 - π)}^{2}} + \frac{1}{{(6 + π)}^{2}} + \frac{1}{{(10 - π)}^{2}} + \frac{1}{{(10 + π)}^{2}} + . . = \frac{π^{2}}{16} {s e c}^{2} (\frac{π^{2}}{4})$ $P r o v e o r d i s p r o v e$
	Answered by mindispower last updated on 27/Jan/21
	$S=Σ_(k≥0) (1/((4k+2−π)^2 ))+(1/((4k+2+π)^2 ))=(1/(16))Σ_(k≥0) (1/((k+((2−π)/4))^2 ))+(1/((k+((2+π)/4))^2 )) cot(x)=(1/x)+Σ_(n≥1) (1/((x−πn)))+(1/((x+πn))) ⇒tg(πx)=cot((π/2)−x)=(2/(π−2x))+Σ_(n≥1) (2/(π−2x−2πn))+(2/(π−2x+2πn)) ⇒1+tg^2 (x)=(4/((π−2x)^2 ))+Σ_(n≥1) (4/((π−2x−2πn)^2 ))+(4/((π−2x+2πn)^2 ))=H x=(π^2 /4) ⇒H=(4/((π−(π^2 /2))^2 ))+Σ_(n≥1) (4/((π−(π^2 /2)−2πn)))+(4/((π−(π^2 /2)+2πn)^2 ))=A+B B=Σ_(n≥0) {(4/((−π−(π^2 /2)−2nπ)^2 ))+(4/((π+2πn−(π^2 /2))^2 ))}−(4/((π−(π^2 /2))^2 )) =(1/π^2 )Σ_(n≥0) (1/((((2+π)/4)+n)^2 ))+(1/((n+((2−π)/4))^2 ))−(4/((π−(π^2 /2)))) =((16)/π^2 )S−(4/((π−(π^2 /2))^2 ))=sec^2 ((π^2 /4))−(4/((π−(π^2 /2))^2 )) ⇒S=(π^2 /(16))sec^2 ((π^2 /4))$
	$S = \sum_{k ⩾ 0} \frac{1}{{(4 k + 2 - π)}^{2}} + \frac{1}{{(4 k + 2 + π)}^{2}} = \frac{1}{16} \sum_{k ⩾ 0} \frac{1}{{(k + \frac{2 - π}{4})}^{2}} + \frac{1}{{(k + \frac{2 + π}{4})}^{2}}$ $c o t (x) = \frac{1}{x} + \sum_{n ⩾ 1} \frac{1}{(x - π n)} + \frac{1}{(x + π n)}$ $\Rightarrow t g (π x) = c o t (\frac{π}{2} - x) = \frac{2}{π - 2 x} + \sum_{n ⩾ 1} \frac{2}{π - 2 x - 2 π n} + \frac{2}{π - 2 x + 2 π n}$ $\Rightarrow 1 + {t g}^{2} (x) = \frac{4}{{(π - 2 x)}^{2}} + \sum_{n ⩾ 1} \frac{4}{{(π - 2 x - 2 π n)}^{2}} + \frac{4}{{(π - 2 x + 2 π n)}^{2}} = H$ $x = \frac{π^{2}}{4}$ $\Rightarrow H = \frac{4}{{(π - \frac{π^{2}}{2})}^{2}} + \sum_{n ⩾ 1} \frac{4}{(π - \frac{π^{2}}{2} - 2 π n)} + \frac{4}{{(π - \frac{π^{2}}{2} + 2 π n)}^{2}} = A + B$ $B = \sum_{n ⩾ 0} {\frac{4}{{(- π - \frac{π^{2}}{2} - 2 n π)}^{2}} + \frac{4}{{(π + 2 π n - \frac{π^{2}}{2})}^{2}}} - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}}$ $= \frac{1}{π^{2}} \sum_{n ⩾ 0} \frac{1}{{(\frac{2 + π}{4} + n)}^{2}} + \frac{1}{{(n + \frac{2 - π}{4})}^{2}} - \frac{4}{(π - \frac{π^{2}}{2})}$ $= \frac{16}{π^{2}} S - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}} = {s e c}^{2} (\frac{π^{2}}{4}) - \frac{4}{{(π - \frac{π^{2}}{2})}^{2}}$ $\Rightarrow S = \frac{π^{2}}{16} {s e c}^{2} (\frac{π^{2}}{4})$
	Commented by Dwaipayan Shikari last updated on 27/Jan/21

	$G r e a t s i r! I h a d p o s t e d t h i s p r o b l e m o n B r i l l i a n t . Y o u c o u l d$ $c h e c k t h e i r a l s o$
	Commented by mindispower last updated on 27/Jan/21

	$w i t h e p l e a s u r s i r i h a v e n o t y e t B r i l l i a n t$
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