Calculate summation of square roots i.e $$\sum_{i=1}^N\sqrt{i}$$
I tried to search for its formula on the net but I couldn't find any of its sources.
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These are generalized harmonic numbers $H_N^{\left(-\frac{1}{2}\right)}$ – Claude Leibovici Sep 19 '14 at 19:36
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@ClaudeLeibovici :How to calculate them accurately for large numbers? – typedef Sep 19 '14 at 19:42
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the easiest way is to use the Euler-Maclaurin formula for integrals and sums or just compare to the corresponding integral. There's not 'closed-form solution'. – Alex Sep 19 '14 at 21:00
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As I said in a comment $$\sum_{i=1}^N \sqrt i=H_N^{\left(-\frac{1}{2}\right)}$$ where $H$ represents a generalized harmonic number.
For large values of $N$, the following approximation is quite interesting and extremely accurate $$\sum_{i=1}^N \sqrt i=\frac{2 N^{3/2}}{3}+\frac{\sqrt{N}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{\sqrt{\frac{1}{N}}}{24}+O\left(\left(\frac{1}{N} \right)^2\right)$$ and very accurate; for example, for $N=10$, the value is $22.46827819$ while the truncated approximation lead to $22.46827983$; for even $N=2$, the value is $2.414213562$ for an approximation of $2.414301422$
Claude Leibovici
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How to calculate them accurately for large numbers?
$$\frac23N\sqrt{N}\leqslant\sum_{i=1}^N\sqrt{i}\leqslant\frac23(N+1)\sqrt{N+1}-\frac23$$ Both bounds are $$\frac23N\sqrt{N}+O(\sqrt{N}).$$
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How might you give a similar expression for $\sum_{i=1}^N i^{2/3}$? $\sum_{i=1}^N \sqrt{i}\log^2 i$? – EGME May 09 '22 at 13:09