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Question Number 144500 by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{calculate}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2+4}$$

Answered by Dwaipayan Shikari last updated on 26/Jun/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+a^2}=\frac{\pi }{2a}coth\left(\frac{\pi }{a}\right)-\frac{1}{2a^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+4}=\frac{\pi }{4}coth\left(\frac{\pi }{2}\right)-\frac{1}{8}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+4}=\frac{\pi }{4}\coth \left(\frac{\pi }{2}\right)+\frac{7}{8}$$

Answered by ArielVyny last updated on 25/Jun/21

$$wehavef\left(t\right)=\frac{1}{t^2+4}positivanddecreasing$$$$in[0.\mathrm{\infty }]then$$$${\int }_0^{\mathrm{\infty }}\frac{1}{t^2+4}dt={\int }_0^{\mathrm{\infty }}\frac{1}{4(1+\frac{t^2}{4})}dt=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{1}{1+\frac{t^2}{4}}dt$$$$u=\frac{t}{4}du=\frac{1}{4}dt$$$${\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}du={\left[arctg\right]}_0^{\mathrm{\infty }}=\frac{\pi }{2}$$$$$$

Commented by mathmax by abdo last updated on 26/Jun/21

$$\mathrm{the}\mathrm{Q}\mathrm{is}\mathrm{a}\mathrm{sum}\mathrm{not}\mathrm{integral}\mathrm{sir}\mathrm{viny}...$$

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