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I was interested in estimating the sum of the form $$ \sum_{j=1}^{N} \{ \sqrt{j} \}. $$ I was wondering if there is a reference or maybe some one could help me figure out what to do. Thanks! $\{ \alpha\}$ denotes the fractional part of the real number $\alpha$.

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Naively, $\lfloor \sqrt{N}\rfloor$ of the integers from $1$ to $N$ are perfect squares, so an upper bound is $N-\lfloor \sqrt{N}\rfloor$. –  vadim123 Oct 9 '13 at 15:28

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up vote 3 down vote accepted

The desired sum is equal to $$\left(\sum_{j=1}^N \sqrt{j}\right) - \left(\sum_{j=1}^N \lfloor \sqrt{j}\rfloor\right)$$

The second sum is computed here as $(N+1)a - \frac{a^3}{3} - \frac{a^2}{2} - \frac{a}{6}$, where $a=\lfloor \sqrt{N+1}\rfloor$.

The first sum may be estimated by an integral as follows $$\frac{2}{3}N^{3/2}=\int_0^N\sqrt{x}dx < \left(\sum_{j=1}^N \sqrt{j}\right) < \int_1^{N+1} \sqrt{x}dx=\frac{2}{3}((N+1)^{3/2}-1)$$

The difference between the upper and lower estimates is $O(\sqrt{N})$, which isn't too bad I think. With care the integrals can be refined, sharpening the estimates.

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Thank you very much! –  Tom Mosher Oct 9 '13 at 16:26
    
My pleasure. $~~$ –  vadim123 Oct 9 '13 at 16:38

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