TITLE: Sum converging to $\zeta(\tfrac12) + 2$
QUESTION [5 upvotes]: In answering this question, I obtained the infinite sum $$\sum _{k=1}^{\infty } \frac{\left(\sqrt{k+1}-\sqrt{k}\right)^2}{\sqrt{k}}.$$
To prove convergence, we don't actually have to figure out what this sum is; just that it is bounded above by, for instance, $\sum_{k=1}^\infty \frac1{4k^{3/2}}$. However, Mathematica tells me that this sum does actually have a relatively simple closed form:
$$\sum _{k=1}^{\infty } \frac{\left(\sqrt{k+1}-\sqrt{k}\right)^2}{\sqrt{k}} = 2 + \zeta(\tfrac12).$$
If this answer had $\zeta(s)$ for $s>1$ in it, I would expect it to be possible to fiddle with the sum until the summation for $\zeta(s)$ popped out, but as it is, I'm lost. How can we obtain this result?
Also, does it generalize in some way, for instance to an expression with $\sqrt[3]{k}$ and $\zeta(\frac13)$?

REPLY [5 votes]: \begin{align*}
  \frac{(\sqrt{k+1}-\sqrt{k})^2}{\sqrt{k}} &=
  \frac{k+1-2\sqrt{k}\sqrt{k+1}+k}{\sqrt{k}} \\
  &= 2(\sqrt{k}-\sqrt{k+1})+\frac{1}{\sqrt{k}} \\
  \sum_{k=1}^{N} \frac{(\sqrt{k+1}-\sqrt{k})^2}{\sqrt{k}}
  &= \sum_{k=1}^{N} \frac{1}{\sqrt{k}}-2\sqrt{N+1}+2 \\
  &= \sum_{k=1}^{N} \frac{1}{\sqrt{k}}-2
     \left[
       \sqrt{N}+O\left( \frac{1}{\sqrt{N}} \right)
     \right]+2 \\
  \lim_{N\to \infty} \sum_{k=1}^{N} \frac{(\sqrt{k+1}-\sqrt{k})^2}{\sqrt{k}}
  &= \zeta \left( \frac{1}{2} \right)+2
\end{align*}

Note that $$
\lim_{n\to \infty}
\left(
  1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}-2\sqrt{n}
\right)=
\zeta \left( \frac{1}{2} \right)=
-1.4603545088 \ldots$$

See another answer here.
For analytic continuation of Riemannn zeta function see the link here.