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

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

Answered by mindispower last updated on 24/Feb/21

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

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

$$Greatsir!Ihavealsofoundso...$$

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