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

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

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

$$Ihavetriedthisapproximation$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}f\left(n\right)={\int }_0^{\mathrm{\infty }}f\left(x\right)dx+\underset{z\to \mathrm{\infty }}{\lim }\frac{f\left(z\right)+f\left(1\right)}{2}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\beta }_{2k}}{\left(2k\right)!}(f^{(2k-1)}\left(z\right)-f^{2k-1}\left(0\right))$$$$\mathcal{E}\mathit{u}\mathit{l}\mathit{e}\mathit{r}\mathfrak{M}\mathit{a}\mathit{c}\mathit{l}\mathit{a}\mathit{u}\mathit{r}\mathit{i}\mathit{n}\mathit{s}\mathit{u}\mathit{m}$$$$\mathit{f}\left(\mathit{z}\right)=\frac{\sqrt{\mathit{z}}}{{\mathit{z}}^2+1}$$$$\sum _{n⩾0}\mathit{f}\left(\mathit{z}\right)={\int }_0^{\mathrm{\infty }}\frac{\sqrt{\mathit{z}}}{{\mathit{z}}^2+1}\mathit{d}\mathit{z}+\frac{1}{4}+\underset{\vartheta \to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\beta }_{2k}}{\left(2k\right)!}(f^{2k-1}\left(\vartheta \right)-f^{2k-1}\left(0\right))$$$$=\frac{\pi }{2\sqrt{2}}+\frac{1}{4}\pm \left\{\left(\stackrel{\mathrm{\Phi }}{\mathrm{\Lambda }}\right)\right\}\to Hardertoapproximate$$$$\stackrel{\mathrm{\Phi }}{\mathrm{\Lambda }}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{\beta }_{2k}}{\left(2k\right)!}{\left(\left(\frac{{\partial }^{2k-1}\left(\frac{\sqrt{z}}{z^2+1}\right)}{\partial z^{2k-1}}\right)\right)}_0^{\mathrm{\infty }}$$

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

$$Anybetterwaysirs?$$

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