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This question already has an answer here:

What's the the result of:

$$\sum_{k=1}^{n}{\sqrt{k}+1}$$

Thanks.

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marked as duplicate by user99914, Harish Chandra Rajpoot, Paul, Martin Sleziak, Community Oct 18 '15 at 11:21

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  • $\begingroup$ I think there is no closed form, but you can find an asymptotic expansion easily with an integral comparaison $\endgroup$ – M.LTA Oct 18 '15 at 10:13
  • $\begingroup$ Why do you expect a closed form of this series? $\endgroup$ – Dontknowanything Oct 18 '15 at 10:17
  • $\begingroup$ ramanujan.sirinudi.org/Volumes/published/ram09.pdf. This question has been asked before, I cant find it. $\endgroup$ – Aditya Agarwal Oct 18 '15 at 10:19
  • $\begingroup$ arxiv.org/pdf/1204.0877.pdf $\endgroup$ – Aditya Agarwal Oct 18 '15 at 10:19
  • $\begingroup$ I expect as @M.LTA says an integral comparison. I know that it's divergent, but want an approximation (for example to know sum to n=100, n=200...) $\endgroup$ – Santiago Gil Oct 18 '15 at 10:20
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I will assume we want an approximation of $S_n=\sum_{k=1}^n \sqrt{k}$.

First, observe that $\sqrt{k} \leqslant \int_{k}^{k+1} \sqrt{x} \; \mathrm{d}x \leqslant \sqrt{k+1}$ so that : $$ \int_{k-1}^{k} \sqrt{x} \; \mathrm{d}x \leqslant \sqrt{k} \leqslant \int_{k}^{k+1} \sqrt{x} \; \mathrm{d}x $$ by summation we have : $$\int_{0}^{n} \sqrt{x} \; \mathrm{d}x \leqslant \sum_{k=1}^{n+1} \sqrt{k} \leqslant \int_{1}^n \sqrt{x} \; \mathrm{d}x$$ wich lead to :

$$\frac{2n^{3/2}}{3} \leqslant \sum_{k=1}^{n+1} \sqrt{k} \leqslant \frac{2\left((n+1)^{3/2}-1\right)}{3} $$

You can continue further if you want, always the same method.

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  • $\begingroup$ Thanks, it's what I searched. Is there another approximation to k = i (this means, the serie doesn't begin with 1). $\endgroup$ – Santiago Gil Oct 18 '15 at 10:53
  • $\begingroup$ Yes of course, you just take the summation from i to n (when I said "by summation we have") $\endgroup$ – M.LTA Oct 18 '15 at 10:56

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