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Question Number 118768 by mathdave last updated on 19/Oct/20

Answered by mindispower last updated on 23/Oct/20

$$\underset{m=0}{\sum ^n}{e}^{imx}=\frac{1-{\left(e^{ix}\right)}^{n+1}}{1-e^{ix}}=\frac{e^{i\frac{nx}{2}}(e^{-i\frac{(n+1)x}{2}}-e^{i\frac{(n+1)x}{2}})}{e^{-i\frac{x}{2}}-e^{i\frac{x}{2}}}$$$$=e^{i\frac{nx}{2}}.\frac{sin\left(\frac{n+1}{2}x\right)}{sin\left(\frac{x}{2}\right)}$$$$\Rightarrow \frac{sin\left(\frac{2n+1}{2}x\right)}{sin\left(\frac{x}{2}\right)}=e^{-inx}\underset{m=0}{\sum ^{2n}}{e}^{imx}$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ikx}}{1+x^2}dx,k⩾0$$$$C_R=\{{Re}^{ia},a\in [0,\pi ]\}$$$${\int }_{C_R}\frac{e^{ikx}}{1+x^2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ikx}}{1+x^2}dx+{\int }_0^{\pi }\frac{{iRe}^{-ksin\left(x\right)}dx}{1+R^2{e}^{2ix}}$$$$=2i\pi .\frac{e^{-k}}{2i}=\frac{\pi }{e^k},$$$$\Rightarrow \sum _{m⩽2n}\frac{\pi }{e^{\mid m-n\mid }}=\sum _{m⩽n}\frac{\pi }{e^{n-m}}+\sum _{m>n}\frac{\pi }{e^{m-n}}$$$$=\frac{\pi }{e^n}\left(\frac{1-e^{(n+1)}}{1-e}\right)+\pi e^n\left(e^{-(n+1)}\left(\frac{1-e^{-\left(n\right)}}{1-e^{-1}}\right)\right)$$$$limn\to 0$$$$=-\frac{\pi e}{1-e}+\pi e^{-1}.\frac{1}{1-e^{-}}$$$$=\frac{\pi }{e-1}(e+1)=\pi \frac{\frac{e^{\frac{1}{2}}}{2}+\frac{e^{-\frac{1}{2}}}{2}}{\frac{1}{2}(e^{\frac{1}{2}}-e^{-\frac{1}{2}})}=\pi coth\left(\frac{1}{2}\right)$$$$$$$$$$$$$$

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