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I have the following sum $\sum\limits_{i=1}^{\lceil \sqrt n \rceil}\sqrt i$.

This sum is $\leq n$, but what is a good bound and what is the method to compute this type of sums?

Dingo13
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    There is no general closed formula for these sums. (But what is a sum up to $\sqrt n$ when $\sqrt n$ is not an integer?) – Did Jan 30 '19 at 10:08
  • @Did Thank you Did. I put the ceiling function. – Dingo13 Jan 30 '19 at 10:10
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    "Good bounds": $$\int_0^{\lceil\sqrt n\rceil}\sqrt xdx\leqslant S_n\leqslant\int_1^{\lceil\sqrt n\rceil+1}\sqrt xdx$$ hence $$\frac23\lceil\sqrt n\rceil\sqrt{\lceil\sqrt n\rceil}\leqslant S_n\leqslant\frac23(\lceil\sqrt n\rceil+1)\left(\sqrt{\lceil\sqrt n\rceil+1}-1\right)$$ which can be simplified into $$\frac23n^{3/4}\leqslant S_n\leqslant\frac23(\sqrt n+2)^{3/2}$$ in particular, $$S_n=\frac23n^{3/4}+O(n^{1/4})$$ – Did Jan 30 '19 at 10:19
  • @Did Thank you very much for your detailed comment. – Dingo13 Jan 30 '19 at 10:21
  • @Did, please make that comment an answer. – lhf Jan 30 '19 at 10:53

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Assuming that the upper limit for the sum is $m$, then $$ \sum\limits_{i=1}^{m}\sqrt i \le \int_1^{m+1} \! \! \sqrt{x} \, dx = \frac23 ((m + 1)^{3/2} - 1) $$

lhf
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  • Thank you lhf. So a bound for the sum when $m=\sqrt{n}$ is $c \cdot n^{3/4}$ for a constant $c$? – Dingo13 Jan 30 '19 at 10:16