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Question Number 133708 by Dwaipayan Shikari last updated on 23/Feb/21

Σ_(n=1) ^∞ ((cos((π+e)n))/n^4 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\left(\pi+{e}\right){n}\right)}{{n}^{\mathrm{4}} } \\ $$

Answered by mindispower last updated on 24/Feb/21

Σ_(n≥1) ((sin(nx))/n)=Im Σ_(n≥1) (e^(inx) /n)=Im(−ln(1−e^(ix) )) −arctan(((−sin(x))/(1−cos(x))))) =−(arctan(((1−cos(x))/(−sin(x)))−(π/2))=((π−x)/2) ⇒Σ_(n≥1) ((−cos(nx))/n^2 )=(π/2)x−(x^2 /4)+c ⇒−Σ((cos(nx))/n^2 )=((πx)/2)−(x^2 /4)−(π^2 /6) −Σ_(n≥1) ((sin(nx))/n^3 )=((πx^2 )/4)−(x^3 /(12))−(π^2 /6)x ⇒Σ_(n≥1) ((cos(nx))/n^4 )=((πx^3 )/(12))−(x^4 /(48))−((π^2 x^2 )/(12))+c,x=0⇒ c=ζ(4) x=π+e Σ_(n≥1) ((cos(π+e)n)/n^4 )=(π/(12))(π+e)^3 −(((π+e)^4 )/(48))−(π^2 /(12))(π+e)^2 +ζ(4)

$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{n}}={Im}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{e}^{{inx}} }{{n}}={Im}\left(−{ln}\left(\mathrm{1}−{e}^{{ix}} \right)\right) \\ $$$$\left.−{arctan}\left(\frac{−{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\right)\right) \\ $$$$=−\left({arctan}\left(\frac{\mathrm{1}−{cos}\left({x}\right)}{−{sin}\left({x}\right)}−\frac{\pi}{\mathrm{2}}\right)=\frac{\pi−{x}}{\mathrm{2}}\right. \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{−{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{c} \\ $$$$\Rightarrow−\Sigma\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=\frac{\pi{x}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{3}} }=\frac{\pi{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{3}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{x} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{4}} }=\frac{\pi{x}^{\mathrm{3}} }{\mathrm{12}}−\frac{{x}^{\mathrm{4}} }{\mathrm{48}}−\frac{\pi^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{12}}+{c},{x}=\mathrm{0}\Rightarrow \\ $$$${c}=\zeta\left(\mathrm{4}\right) \\ $$$${x}=\pi+{e} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\pi+{e}\right){n}}{{n}^{\mathrm{4}} }=\frac{\pi}{\mathrm{12}}\left(\pi+{e}\right)^{\mathrm{3}} −\frac{\left(\pi+{e}\right)^{\mathrm{4}} }{\mathrm{48}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\left(\pi+{e}\right)^{\mathrm{2}} +\zeta\left(\mathrm{4}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 24/Feb/21

Great sir! I have also found so...

$${Great}\:{sir}!\:{I}\:{have}\:{also}\:{found}\:{so}... \\ $$

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