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Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that $$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.

With some more calculations, we get $$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + \frac{1}{2} (k^{1/2}-(k-1)^{-3/2}) + O(k^{-1/2}),$$ hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + \frac{1}{2} n^{1/2} + C + O(n^{1/2})$ for some constant $C$, because $\sum_n^\infty O(k^{-3/2}) = O(n^{-1/2})$.

Now let's go further. I have made the following calculation $$k^{1/2} = \frac{3}{2} \Delta_{3/2}(k) + \frac{1}{2} \Delta_{1/2}(k) + \frac{1}{24} \Delta_{-1/2}(k) + O(k^{-5/2}),$$ where $\Delta_\alpha(k) = k^\alpha-(k-1)^{\alpha}$. Hence : $$\sum_{k=1}^n \sqrt{k} = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2}).$$ And one can continue ad vitam aeternam, but the only term I don't know how to compute is the constant term.

How do we find $C$ ?

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    $\begingroup$ you want to calculate the taylor series of a sequence? for taylor you need a differentiable function ... $\endgroup$
    – Dominic Michaelis
    Commented Mar 10, 2013 at 17:44
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    $\begingroup$ The $\sum_{1 \le k \le n} O(k^{-1/2}) = O(n^{-1/2})$ gives me the willies... to start with, the largest term of the sum $\sum_{1 \le k \le n} k^{-1/2}$ is $O(1)$. If anything, express the missing terms more precisely, and estimate (or bound) $\sum_{1 \le k \le n} f(k)$ with $\int_1^n f(x) dx$. Or use Euler-Maclaurin's formula directly $\endgroup$
    – vonbrand
    Commented Mar 10, 2013 at 18:32
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    $\begingroup$ @vonbrand. I have written $O(n^{+1/2})$ not $O(n^{-1/2})$. Euler-Maclaurin tells : $\int_1^n \sqrt{t} dt = \frac{1}{2}\sqrt{1} + \dots + \sqrt{1/2} \sqrt{n} + \frac{B_2}{2}(\frac{1}{\sqrt{n}}-\frac{1}{2}) + \int_1^n \frac{P_2(t)}{2} \frac{-1/4}{t^{3/2}} dt$. But $\int_1^n \frac{P_2(t)}{2} \frac{-1/4}{t^{3/2}} dt = Constant + O(1/n^{1/2})$, so this method doesn't give $C$ (or am I misusing it ?) $\endgroup$
    – Auguste Hoang Duc
    Commented Mar 10, 2013 at 20:15
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    $\begingroup$ $C\ne 0$. $C=\zeta(-\frac 12)\approx-0.207886224977$. $\endgroup$
    – David Moews
    Commented Mar 11, 2013 at 3:02
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    $\begingroup$ A calculation of the complete asymptotic expansion by Mellin transforms can be found here. $\endgroup$
    – Marko Riedel
    Commented Jul 17, 2013 at 4:04

3 Answers 3

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Let us substitute into the sum $$\sqrt k=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{k e^{-kx}dx}{\sqrt x}. $$ Exchanging the order of summation and integration and summing the derivative of geometric series, we get \begin{align*} \mathcal S_N:= \sum_{k=1}^{N}\sqrt k&=\frac{1}{\sqrt \pi }\int_0^{\infty}\frac{\left(e^x-e^{-(N-1)x}\right)-N\left(e^{-(N-1)x}-e^{-Nx}\right)}{\left(e^x-1\right)^2}\frac{dx}{\sqrt x}=\\&=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}\right)\frac{dx}{x\sqrt x}=\\ &=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}\right)\frac{dx}{x\sqrt x}. \end{align*} To extract the asymptotics of the above integral it suffices to slightly elaborate the method used to answer this question. Namely \begin{align*} \mathcal S_N&=\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\frac{1-e^{-Nx}}{e^x-1}+\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)-\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)\right)\frac{dx}{x\sqrt x}=\\ &={\color{red}{\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(1-e^{-Nx}\right)\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}}}+\\&+ {\color{blue}{\frac{1}{2\sqrt\pi}\int_0^{\infty} \left(N-\left(1-e^{-Nx}\right)\left(\frac1x-\frac12\right)\right)\frac{dx}{x\sqrt x}}}. \end{align*} The reason to decompose $\mathcal S_N$ in this way is that

  • the red integral has an easily computable finite limit: since $\frac1x-\frac12-\frac{1}{e^x-1}=O(x)$ as $x\to 0$, we can simply neglect the exponential $e^{-Nx}$.

  • the blue integral can be computed exactly.

Therefore, as $N\to \infty$, we have $$\mathcal S_N={\color{blue}{\frac{\left(4n+3\right)\sqrt n}{6}}}+ {\color{red}{\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}+o(1)}},$$ and the finite part you are looking for is given by $$C=\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}=\zeta\left(-\frac12\right).$$

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As shown in this answer, we can use the Euler-Maclaurin Sum Formula to get $$ \sum_{k=1}^n\sqrt{k}=\frac23n^{3/2}+\frac12n^{1/2}+\zeta\left(-\frac12\right)+\frac1{24}n^{-1/2}-\frac1{1920}n^{-5/2}+\frac1{9216}n^{-9/2}+O\left(n^{-13/2}\right) $$

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With Mathematica it very easy:

Series[Sum[Sqrt[k], {k, 1, n}], {n, Infinity, 2}]// TeXForm

$$\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{\sqrt{\frac{1}{n}}}{24}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$

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