# Sum of finite series involving square roots [duplicate]

What's the the result of:

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

Thanks.

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

• I think there is no closed form, but you can find an asymptotic expansion easily with an integral comparaison – M.LTA Oct 18 '15 at 10:13
• Why do you expect a closed form of this series? – Dontknowanything Oct 18 '15 at 10:17
• ramanujan.sirinudi.org/Volumes/published/ram09.pdf. This question has been asked before, I cant find it. – Aditya Agarwal Oct 18 '15 at 10:19
• arxiv.org/pdf/1204.0877.pdf – Aditya Agarwal Oct 18 '15 at 10:19
• 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...) – Santiago Gil Oct 18 '15 at 10:20

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