Question Number 197581 by mokys last updated on 22/Sep/23
$${find}\:\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\sqrt{{n}}\:? \\ $$
Answered by salvatore last updated on 07/Jan/24
![w(1)=(1/(24)) w(m)= (((1/2)),(( m)) ) (B_(m+1) /(m+1)) −(−1)^m Σ_(h=1) ^(m−1) (−1)^h { (m),(h) :}}w(h) Σ_(k=1) ^n (√k) =−((ζ((3/2)))/(4π)) +(√n) [(2/3) n +(1/2) +Σ_(k=1) ^∞ ((w(k))/((n+1)_((k)) ))] B_h hesimo numero di Bernoulli](Q197611.png)
$$\left.{w}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{24}}\:\:\:{w}\left({m}\right)=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\:{m}}\end{pmatrix}\:\frac{{B}_{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\:−\left(−\mathrm{1}\right)^{{m}} \underset{{h}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}\:\left(−\mathrm{1}\right)^{{h}} \begin{cases}{{m}}\\{{h}}\end{cases}\right\}{w}\left({h}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{{k}}\:\:=−\frac{\zeta\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{4}\pi}\:+\sqrt{{n}}\:\left[\frac{\mathrm{2}}{\mathrm{3}}\:{n}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{w}\left({k}\right)}{\left({n}+\mathrm{1}\right)_{\left({k}\right)} }\right] \\ $$$${B}_{{h}} \:{hesimo}\:{numero}\:{di}\:{Bernoulli} \\ $$