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Question Number 128931 by Dwaipayan Shikari last updated on 11/Jan/21

Approximate Σ_(n=1) ^∞ ((√n)/(n^2 +1))

$${Approximate} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\sqrt{{n}}}{{n}^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by Dwaipayan Shikari last updated on 11/Jan/21

I have tried this approximation Σ_(n=1) ^∞ f(n)=∫_0 ^∞ f(x)dx+lim_(z→∞) ((f(z)+f(1))/2)+Σ_(k=1) ^∞ (β_(2k) /((2k)!))(f^((2k−1)) (z)−f^(2k−1) (0)) Euler Maclaurin sum f(z)=((√z)/(z^2 +1)) Σ_(n≥0) f(z)=∫_0 ^∞ ((√z)/(z^2 +1))dz +(1/4)+lim_(ϑ→∞) Σ_(k=1) ^∞ (β_(2k) /((2k)!))(f^(2k−1) (ϑ)−f^(2k−1) (0)) = (π/(2(√2)))+(1/4)±{(Λ^Φ ) }→Harder to approximate Λ^Φ =Σ_(k=1) ^∞ (β_(2k) /((2k)!))((((∂^(2k−1) (((√z)/(z^2 +1))) )/∂z^(2k−1) )))_0 ^∞

$${I}\:{have}\:{tried}\:{this}\:{approximation} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{f}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right){dx}+\underset{{z}\rightarrow\infty} {\mathrm{lim}}\frac{{f}\left({z}\right)+{f}\left(\mathrm{1}\right)}{\mathrm{2}}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta_{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!}\left({f}^{\left(\mathrm{2}{k}−\mathrm{1}\right)} \left({z}\right)−{f}^{\mathrm{2}{k}−\mathrm{1}} \left(\mathrm{0}\right)\right) \\ $$$$\boldsymbol{\mathcal{E}{uler}}\:\boldsymbol{\mathfrak{M}{aclaurin}}\:\boldsymbol{{sum}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)=\frac{\sqrt{\boldsymbol{{z}}}}{\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{\boldsymbol{{z}}}}{\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dz}}\:+\frac{\mathrm{1}}{\mathrm{4}}+\underset{\vartheta\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta_{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!}\left({f}^{\mathrm{2}{k}−\mathrm{1}} \left(\vartheta\right)−{f}^{\mathrm{2}{k}−\mathrm{1}} \left(\mathrm{0}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{4}}\pm\left\{\left(\overset{\Phi} {\Lambda}\right)\:\right\}\rightarrow{Harder}\:{to}\:{approximate} \\ $$$$\overset{\Phi} {\Lambda}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta_{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!}\left(\left(\frac{\partial^{\mathrm{2}{k}−\mathrm{1}} \left(\frac{\sqrt{{z}}}{{z}^{\mathrm{2}} +\mathrm{1}}\right)\:}{\partial{z}^{\mathrm{2}{k}−\mathrm{1}} }\right)\right)_{\mathrm{0}} ^{\infty} \\ $$

Commented by Dwaipayan Shikari last updated on 11/Jan/21

Any better way sirs?

$${Any}\:{better}\:{way}\:\:{sirs}? \\ $$

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