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Question Number 176566 by mnjuly1970 last updated on 21/Sep/22

$$----$$$$calculate:\mathrm{\Phi }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+1).e^{4n+2}}=?$$$$where{}^{″}e{}^{″}iseulernumber.$$$$\prec solution\succ$$$$\mathrm{\Phi }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{e^{4n+2}}{\int }_0^1{x}^{2n}dx$$$$=\frac{1}{e^2}{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{x^2}{e^4}\right)}^{n}dx$$$$=\frac{1}{e^2}{\int }_0^1\frac{1}{1-{\left(\frac{x}{e^2}\right)}^2}dx=\frac{1}{2e^2}{\int }_0^1\frac{1}{1-\frac{x}{e^2}}+\frac{1}{1+\frac{x}{e^2}}dx$$$$=\frac{1}{2}ln\left(\frac{1+\frac{1}{e^2}}{1-\frac{1}{e^2}}\right)={tanh}^{-1}\left(\frac{1}{e^2}\right)$$$$\therefore \mathrm{\Phi }={coth}^{-1}\left(e^2\right)◼m.n$$$$$$

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