Approximate-n-1-n-n-2-1-

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

A p p r o x i m a t e

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

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

I h a v e t r i e d t h i s a p p r o x i m a t i o n

\underset{n = 1}{\sum^{\infty}} f (n) = \int_{0}^{\infty} f (x) d x + \lim_{z \to \infty} \frac{f (z) + f (1)}{2} + \underset{k = 1}{\sum^{\infty}} \frac{β_{2 k}}{(2 k)!} (f^{(2 k - 1)} (z) - f^{2 k - 1} (0))

E u l e r M a c l a u r i n s u m

f (z) = \frac{\sqrt{z}}{z^{2} + 1}

\sum_{n ⩾ 0} f (z) = \int_{0}^{\infty} \frac{\sqrt{z}}{z^{2} + 1} d z + \frac{1}{4} + \lim_{ϑ \to \infty} \underset{k = 1}{\sum^{\infty}} \frac{β_{2 k}}{(2 k)!} (f^{2 k - 1} (ϑ) - f^{2 k - 1} (0))

= \frac{π}{2 \sqrt{2}} + \frac{1}{4} \pm {(\overset{Φ}{Λ})} \to H a r d e r t o a p p r o x i m a t e

\overset{Φ}{Λ} = \underset{k = 1}{\sum^{\infty}} \frac{β_{2 k}}{(2 k)!} {((\frac{\partial^{2 k - 1} (\frac{\sqrt{z}}{z^{2} + 1})}{\partial z^{2 k - 1}}))}_{0}^{\infty}

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

A n y b e t t e r w a y s i r s ?