find the value of the following series . Ω= Σ_(n=1) ^∞ (( cos(((nπ)/4) ))/n^( 2) ) =?


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Question Number 191528 by mnjuly1970 last updated on 25/Apr/23

$f i n d t h e v a l u e o f t h e$ $f o l l o w i n g s e r i e s .$ $Ω = \underset{n = 1}{\sum^{\infty}} \frac{c o s (\frac{n π}{4})}{n^{2}} = ?$
	Answered by witcher3 last updated on 25/Apr/23
	$Σ_(n≥1) ((cos(n(π/4)))/n^2 )=(1/2)Σ_(n≥1) (((e^((iπ)/4) )^n +(e^(−((iπ)/4)) )^n )/n^2 ) =(1/2){Σ_(n≥1) (((e^((iπ)/4) )^n )/n^2 )+Σ_(n≥1) (((e^(−((iπ)/4)) )^n )/n^2 )} =(1/2)(Li_2 (e^(i(π/4)) )+Li_2 (e^(−((iπ)/4)) )) Li_2 (z)+Li_2 ((1/z))=−ζ(2)−(1/2)ln^2 (−z) Ω=(1/2)(Li_2 (e^(−((iπ)/4)) )+Li_2 ((1/e^(−((iπ)/4)) ))) =(1/2)(−(𝛑^2 /6)−(1/2)ln^2 (−e^(−((iπ)/4)) ))=(1/2)(−(π^2 /6)+((9π^2 )/(32))) =((11π^2 )/(192))$
	$\sum_{n ⩾ 1} \frac{\cos (n \frac{π}{4})}{n^{2}} = \frac{1}{2} \sum_{n ⩾ 1} \frac{{(e^{\frac{i π}{4}})}^{n} + {(e^{- \frac{i π}{4}})}^{n}}{n^{2}}$ $= \frac{1}{2} {\sum_{n ⩾ 1} \frac{{(e^{\frac{i π}{4}})}^{n}}{n^{2}} + \sum_{n ⩾ 1} \frac{{(e^{- \frac{i π}{4}})}^{n}}{n^{2}}}$ $= \frac{1}{2} ({Li}_{2} (e^{i \frac{π}{4}}) + {Li}_{2} (e^{- \frac{i π}{4}}))$ ${Li}_{2} (z) + {Li}_{2} (\frac{1}{z}) = - ζ (2) - \frac{1}{2} \ln^{2} (- z)$ $Ω = \frac{1}{2} ({Li}_{2} (e^{- \frac{i π}{4}}) + {Li}_{2} (\frac{1}{e^{- \frac{i π}{4}}}))$ $= \frac{1}{2} (- \frac{π^{2}}{6} - \frac{1}{2} \ln^{2} (- e^{- \frac{i π}{4}})) = \frac{1}{2} (- \frac{π^{2}}{6} + \frac{9 π^{2}}{32})$ $= \frac{11 π^{2}}{192}$
	Commented by mnjuly1970 last updated on 25/Apr/23

	$t h a n k s a l o t . . . . s i r$
	Commented by witcher3 last updated on 25/Apr/23

	$withe Pleasur Sir$
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