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Question Number 191528 by mnjuly1970 last updated on 25/Apr/23
find the value of the following series . Ω= Σ_(n=1) ^∞ (( cos(((nπ)/4) ))/n^( 2) ) =?
$$ \\ $$$$\:\:\:\:\:\:{find}\:\:{the}\:\:{value}\:\:{of}\:\:{the} \\ $$$$\:\:\:\:\:\:\:{following}\:\:{series}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{cos}\left(\frac{{n}\pi}{\mathrm{4}}\:\right)}{{n}^{\:\mathrm{2}} }\:=? \\ $$
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))
$$\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{n}\frac{\pi}{\mathrm{4}}\right)}{\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} +\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)+\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right) \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\mathrm{z}\right)+\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{z}}\right)=−\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(−\mathrm{z}\right) \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)+\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{32}}\right) \\ $$$$=\frac{\mathrm{11}\pi^{\mathrm{2}} }{\mathrm{192}} \\ $$
Commented by mnjuly1970 last updated on 25/Apr/23
thanks alot ....sir
$${thanks}\:{alot}\:….{sir} \\ $$
Commented by witcher3 last updated on 25/Apr/23
withe Pleasur Sir
$$\mathrm{withe}\:\mathrm{Pleasur}\:\mathrm{Sir} \\ $$

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