How to calculate the asymptotic expansion of $\sum \sqrt{k}$?

Solution 1:

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 $$C=\frac{1}{2\sqrt\pi}\int_0^{\infty}\left(\frac1x-\frac12-\frac{1}{e^x-1}\right)\frac{dx}{x\sqrt x}=\zeta\left(-\frac12\right).$$

Solution 2:

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