Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 215516 by MrGaster last updated on 09/Jan/25

      prove:        Σ_(n=1) ^∞ (1/(π^2 n^2 +1))=(1/(e^2 −1))

$$\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}} \\ $$

Answered by mr W last updated on 09/Jan/25

Σ_(n=1) ^∞ (1/(π^2 n^2 +1))  =(1/(2i))Σ_(n=1) ^∞ ((1/(πn−i))−(1/(πn+i)))  =(1/(2πi))Σ_(n=1) ^∞ ((1/(n−(i/π)))−(1/(π+(i/π))))  =(1/(2πi))[Σ_(n=1) ^∞ (1/n)−γ−ψ(1−(i/π))−Σ_(n=1) ^∞ (1/n)+γ+ψ(1+(i/π))]  =(1/(2πi))[ψ(1+(i/π))−ψ(1−(i/π))]  =(1/(2πi))[ψ((i/π))+(π/i)−ψ((i/π))−π cot i]  =(1/(2πi))((π/i)−π cot i)  =(1/(2πi))((π/i)−(π/i)coth 1)  =(1/2)(coth 1−1)  =(1/2)(((e+e^(−1) )/(e−e^(−1) ))−1)  =(e^(−1) /(e−e^(−1) ))  =(1/(e^2 −1)) ✓

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\pi^{\mathrm{2}} {n}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\pi{n}−{i}}−\frac{\mathrm{1}}{\pi{n}+{i}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\frac{{i}}{\pi}}−\frac{\mathrm{1}}{\pi+\frac{{i}}{\pi}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\gamma−\psi\left(\mathrm{1}−\frac{{i}}{\pi}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}+\gamma+\psi\left(\mathrm{1}+\frac{{i}}{\pi}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\left[\psi\left(\mathrm{1}+\frac{{i}}{\pi}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\pi}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\left[\psi\left(\frac{{i}}{\pi}\right)+\frac{\pi}{{i}}−\psi\left(\frac{{i}}{\pi}\right)−\pi\:\mathrm{cot}\:{i}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\left(\frac{\pi}{{i}}−\pi\:\mathrm{cot}\:{i}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\left(\frac{\pi}{{i}}−\frac{\pi}{{i}}\mathrm{coth}\:\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{coth}\:\mathrm{1}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{e}+{e}^{−\mathrm{1}} }{{e}−{e}^{−\mathrm{1}} }−\mathrm{1}\right) \\ $$$$=\frac{{e}^{−\mathrm{1}} }{{e}−{e}^{−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}}\:\checkmark \\ $$

Commented by ajfour last updated on 10/Jan/25

https://youtu.be/dvzKfeWOqyw?si=EggqbIQMc8QTN9F6 LCR Series circuit current and voltages discussed.

Commented by MrGaster last updated on 09/Jan/25

Thank you sir.

$$\mathrm{Than}\Bbbk\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com