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Question Number 138163 by mnjuly1970 last updated on 10/Apr/21

              ........nice  ... .... .... calculus.....      prove that::              Ψ=Σ_(n=1) ^∞ ((1/(n^2 π^2 +1)))=^(???) (1/(e^2 −1))          .............

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:........{nice}\:\:...\:....\:....\:{calculus}..... \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}}\right)\overset{???} {=}\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:............. \\ $$

Answered by Dwaipayan Shikari last updated on 10/Apr/21

Σ_(n=1) ^∞ (1/(n^2 +φ^2 ))=(π/(2∅))coth(π∅)−(1/(2∅^2 ))  ⇒Σ_(n=1) ^∞ (1/(n^2 π^2 +1))=(1/π^2 )Σ_(n=1) ^∞ (1/(n^2 +((1/π))^2 ))=(1/π^2 )((π^2 /2)coth(1)−(π^2 /2))  =(1/2)(((e^2 +1)/(e^2 −1))−1)=(1/(e^2 −1))=((1/e)/(e−(1/e)))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\phi^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}\emptyset}{coth}\left(\pi\emptyset\right)−\frac{\mathrm{1}}{\mathrm{2}\emptyset^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\pi}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{coth}\left(\mathrm{1}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{e}^{\mathrm{2}} +\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}\right)=\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}}=\frac{\frac{\mathrm{1}}{{e}}}{{e}−\frac{\mathrm{1}}{{e}}} \\ $$

Commented by mnjuly1970 last updated on 10/Apr/21

    greateful..mr..payan...

$$\:\:\:\:{greateful}..{mr}..{payan}... \\ $$

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