sequences and series - Sum converging to $zeta(tfrac12) + 2$

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 $\zeta(\frac13)$?

Answer

\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.