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Question Number 167820 by MWSuSon last updated on 26/Mar/22
given x^2 =(π^2 /3)+4Σ(−1)^n ((cos(nx))/n^2 ), show that Σ(1/((2n−1)^2 ))=(π^2 /8)
$$\mathrm{given}\:\:\mathrm{x}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} },\:\mathrm{show}\:\mathrm{that}\:\Sigma\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Answered by Mathspace last updated on 26/Mar/22
x^2 =(π^2 /3)+4Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) x=(π/2) ⇒(π^2 /4)=(π^2 /3)+4Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(((nπ)/2)) n=2m⇒cos(n(π/2))=(−1)^m n=2m+1 ⇒cos(((nπ)/2))=0 ⇒ 4Σ_(m=1) ^∞ (((−1)^(2m) (−1)^m )/(4m^2 ))=(π^2 /4)−(π^2 /3) =−(π^2 /(12)) ⇒Σ_(m=1) ^∞ (((−1)^m )/m^2 )=−(π^2 /(12)) ⇒ (1/4)Σ_(n=1) ^∞ (1/n^2 )−Σ_(n=0) ^∞ (1/((2n+1)^2 ))=−(π^2 /(12)) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 ))=(1/4)(π^2 /6)+(π^2 /(12)) =((π^2 +2π^2 )/(24))=((3π^2 )/(24))=(π^2 /8) n=r−1⇒Σ_(r=1) ^∞ (1/((2r−1)^2 ))=(π^2 /8)
$${x}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$${x}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$${n}=\mathrm{2}{m}\Rightarrow{cos}\left({n}\frac{\pi}{\mathrm{2}}\right)=\left(−\mathrm{1}\right)^{{m}} \\ $$$${n}=\mathrm{2}{m}+\mathrm{1}\:\Rightarrow{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{4}\sum_{{m}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{2}{m}} \left(−\mathrm{1}\right)^{{m}} }{\mathrm{4}{m}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow\sum_{{m}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{m}} }{{m}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$=\frac{\pi^{\mathrm{2}} +\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${n}={r}−\mathrm{1}\Rightarrow\sum_{{r}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$

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