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 $\zeta(\frac13)$?


\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.

  • 1
    $\begingroup$ I suggest that you prove that $$\zeta(1/2)=\lim_{N\to \infty }\left( \sum_{k=1}^n\frac{1}{\sqrt{k}}-2\sqrt{N+1}\right)$$ $\endgroup$ – Mark Viola Apr 8 '17 at 4:38
  • $\begingroup$ @Dr.MV, Answer updated, thanks for your comment. $\endgroup$ – Ng Chung Tak Apr 8 '17 at 4:49
  • 2
    $\begingroup$ A better reference on MSE is Rob's post here. -Mark $\endgroup$ – Mark Viola Apr 8 '17 at 5:27

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