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Question Number 67384 by mathmax by abdo last updated on 26/Aug/19

find Σ_(n=1) ^∞ ((cos(nx))/n^2 )

$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} } \\ $$

Commented by mathmax by abdo last updated on 27/Aug/19

we have proved that Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) ⇒ Σ_(n=1) ^∞ (1/n)∫ sin(nx)dx =(1/2)∫(π−x)dx +c ⇒ −Σ_(n=1) ^∞ (1/n^2 )cos(nx) =(1/2){πx−(x^2 /2)} +c =(π/2)x−(x^2 /4) +c ⇒ Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =−(π/2)x +(x^2 /4) −c x=0 ⇒Σ_(n=1) ^∞ (1/n^2 ) =−c ⇒c =−(π^2 /6) ⇒Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =(x^2 /4)−((πx)/2) +(π^2 /6)

$${we}\:{have}\:{proved}\:{that}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\int\:\:{sin}\left({nx}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\pi−{x}\right){dx}\:+{c}\:\Rightarrow \\ $$$$−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }{cos}\left({nx}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\pi{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right\}\:+{c}\:=\frac{\pi}{\mathrm{2}}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+{c}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }\:=−\frac{\pi}{\mathrm{2}}{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:−{c} \\ $$$${x}=\mathrm{0}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=−{c}\:\Rightarrow{c}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi{x}}{\mathrm{2}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

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