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Question Number 132434 by Dwaipayan Shikari last updated on 14/Feb/21

Σ_(n=1) ^∞ ((cos(n))/n^2 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({n}\right)}{{n}^{\mathrm{2}} } \\ $$

Answered by mnjuly1970 last updated on 14/Feb/21

solution: Σ_(n=1) ^∞ ((e^(in) +e^(−in) )/(2n^2 ))=(1/2)[li_2 (e^i )+li_2 (e^(−i) )] =(1/2)(li_2 (e^i )+li_2 ((1/e^i ))) =_(−ζ(2)−(1/2)ln^2 (−x)) ^(li_2 (x)+li_2 ((1/x))) (1/2)(−(π^2 /6)−(1/2)ln^2 (−e^i )) =((−π^2 )/(12)) −(1/4)(2ln(i)+i)^2 =((−π^2 )/(12))−(1/4)(i(π+1)^2 )=((−π^2 )/(12))+(((π+1)^2 )/4)

$$\:\:{solution}: \\ $$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}} +{e}^{−{in}} }{\mathrm{2}{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[{li}_{\mathrm{2}} \left({e}^{{i}} \right)+{li}_{\mathrm{2}} \left({e}^{−{i}} \right)\right] \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({li}_{\mathrm{2}} \left({e}^{{i}} \right)+{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{e}^{{i}} }\right)\right) \\ $$$$\:\:\:\underset{−\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(−{x}\right)} {\overset{{li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)} {=}}\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(−{e}^{{i}} \right)\right) \\ $$$$\:\:\:\: \\ $$$$\:=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{ln}\left({i}\right)+{i}\right)^{\mathrm{2}} \\ $$$$=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}}\left({i}\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$

Commented by Dwaipayan Shikari last updated on 14/Feb/21

Great sir! thanking you..

$${Great}\:{sir}!\:{thanking}\:{you}.. \\ $$

Commented by mnjuly1970 last updated on 14/Feb/21

you are welcome...

$$\:{you}\:{are}\:{welcome}... \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 14/Feb/21

let ϕ(x)=Σ_(n=1) ^∞ ((sin(nx))/n) we have ∫ ϕ(x)dx=−Σ_(n=1) ^∞ ((cos(nx))/n^2 ) ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 )=K−∫ ϕ(x)dx ϕ(x)=Im(Σ_(n=1) ^∞ (e^(inx) /n)) but Σ_(n=1) ^∞ (((e^(ix) )^n )/n) =Ψ(e^(ix) )with Ψ(u) =Σ_(n=1) ^∞ (u^n /n) =−ln(1−u) ⇒Ψ(e^(ix) )=−ln(1−e^(ix) ) =−ln(1−cosx−isinx) =−ln(2sin^2 ((x/2))−2isin((x/2))cos((x/2))) =−ln(−2isin((x/2)){cos((x/2))+isin((x/2))}) =−ln(−2)−ln(i)−ln(sin((x/2))−ln(e^((ix)/2) ) =−ln(e^(iπ) )−ln(2)−ln(e^((iπ)/2) )−ln(sin((x/2)))−((ix)/2) =−iπ−((iπ)/2)−((ix)/2) −ln(sin((x/2)))−ln(2) =i(−((3π)/2)−(x/2))−ln(2sin((x/2))) ⇒ϕ(x)=−(x/2)−((3π)/2) ⇒ ∫ ϕ(x)dx =−(x^2 /4)−((3π)/2)x ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 )=k+(x^2 /4)+((3π)/2)x x=0 ⇒(π^2 /6)=k ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =(x^2 /4)+((3π)/2)x +(π^2 /6) x=1 ⇒Σ_(n=1) ^∞ ((cos(n))/n^2 ) =(1/4)+((3π)/2)+(π^2 /6)

$$\mathrm{let}\:\varphi\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{nx}\right)}{\mathrm{n}}\:\:\mathrm{we}\:\mathrm{have}\:\int\:\varphi\left(\mathrm{x}\right)\mathrm{dx}=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} }=\mathrm{K}−\int\:\varphi\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\varphi\left(\mathrm{x}\right)=\mathrm{Im}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{inx}} }{\mathrm{n}}\right)\:\:\mathrm{but}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{n}} }{\mathrm{n}}\:=\Psi\left(\mathrm{e}^{\mathrm{ix}} \right)\mathrm{with} \\ $$$$\Psi\left(\mathrm{u}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}}\:=−\mathrm{ln}\left(\mathrm{1}−\mathrm{u}\right)\:\Rightarrow\Psi\left(\mathrm{e}^{\mathrm{ix}} \right)=−\mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{\mathrm{ix}} \right) \\ $$$$=−\mathrm{ln}\left(\mathrm{1}−\mathrm{cosx}−\mathrm{isinx}\right)\:=−\mathrm{ln}\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2isin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right) \\ $$$$=−\mathrm{ln}\left(−\mathrm{2isin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left\{\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right\}\right) \\ $$$$=−\mathrm{ln}\left(−\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{i}\right)−\mathrm{ln}\left(\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{e}^{\frac{\mathrm{ix}}{\mathrm{2}}} \right)\right. \\ $$$$=−\mathrm{ln}\left(\mathrm{e}^{\mathrm{i}\pi} \right)−\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)−\mathrm{ln}\left(\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)−\frac{\mathrm{ix}}{\mathrm{2}} \\ $$$$=−\mathrm{i}\pi−\frac{\mathrm{i}\pi}{\mathrm{2}}−\frac{\mathrm{ix}}{\mathrm{2}}\:−\mathrm{ln}\left(\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{i}\left(−\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{ln}\left(\mathrm{2sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\:\Rightarrow\varphi\left(\mathrm{x}\right)=−\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\int\:\varphi\left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{x}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} }=\mathrm{k}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{x} \\ $$$$\mathrm{x}=\mathrm{0}\:\Rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\mathrm{k}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} }\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{x}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{x}=\mathrm{1}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}\right)}{\mathrm{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}\pi}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Commented by Dwaipayan Shikari last updated on 15/Feb/21

Great sir!

$${Great}\:{sir}! \\ $$

Commented by mathmax by abdo last updated on 20/Feb/21

thanks

$$\mathrm{thanks} \\ $$

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