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Question Number 99894 by bachamohamed last updated on 23/Jun/20

     Σ_(n≥0)   (1/(n^2 +1)) = ?

$$\:\:\: \\ $$$$\underset{\boldsymbol{{n}}\geqslant\mathrm{0}} {\sum}\:\:\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}\:=\:? \\ $$

Answered by smridha last updated on 24/Jun/20

s(a)=Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=((𝛑coth(𝛑a))/(2a))−(1/(2a^2 ))     now s(1)=((𝛑coth(𝛑))/2)−(1/2)≈1.0766...  now Σ_(n=0) ^∞ (1/(n^2 +1))=1+Σ_(n=1) ^∞ (1/(n^2 +1))≈2.0766....

$$\boldsymbol{{s}}\left(\boldsymbol{{a}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{a}}^{\mathrm{2}} }=\frac{\boldsymbol{\pi{coth}}\left(\boldsymbol{\pi{a}}\right)}{\mathrm{2}\boldsymbol{{a}}}−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{a}}^{\mathrm{2}} } \\ $$$$\:\:\:\boldsymbol{{now}}\:\boldsymbol{{s}}\left(\mathrm{1}\right)=\frac{\boldsymbol{\pi{coth}}\left(\boldsymbol{\pi}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{0766}... \\ $$$$\boldsymbol{{now}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\approx\mathrm{2}.\mathrm{0766}.... \\ $$

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