Asymptotics of $\sum \sqrt{k}$ and $\sum (-1)^k\sqrt{k}$

Question

I was playing around with series recently and asymptotics of $\sum \sqrt{k}$ and $\sum (-1)^k\sqrt{k}$ were required to solve another problem.

I have dealt with the first one using an integral estimate: $$\forall n\in \mathbb N,\;\; \frac23 n^{3/2}\leq \sum_{k=1}^n\sqrt{k}\leq \frac23 (n+1)^{3/2}$$

Hence $$\sum_{k=1}^n\sqrt{k}=\frac23 n^{3/2} +o(n^{3/2} )$$

I'm clueless with the other one. I can't resort to the same method since the sign keeps changing.

Answer 1

You can use the same technique.

$$ \sum_{k=1}^{2n} (-1)^k \sqrt{k} = \sum_{k=0}^n \sqrt{2k} - \sqrt{2k+1}. $$

The terms of the right-hand summation are positive and decreasing, so you can use the integral test:

$$ \frac23\left(n^{3/2} - (n+1)^{3/2}\right) < \sum_{k=0}^n \sqrt{2k} - \sqrt{2k+1} < \frac23\left((n+1)^{3/2} - (n+2)^{3/2}\right). $$

The number $n^{3/2} - (n+1)^{3/2}$ is somewhere between $\sqrt{n}$ and $\sqrt{n+1}$.

Answer 2

Assume $N$ odd (it is not restrictive). We have$$\sum_{k=1}^{N}\left(-1\right)^{k}\sqrt{k}=\sqrt{2}\sum_{k=1}^{N-1}\sqrt{k}-\sum_{k=1}^{N}\sqrt{2k-1}$$ and the second sum is, by partial summation$$\sum_{k=1}^{N}\sqrt{2k-1}=N\sqrt{2N-1}-\int_{1}^{N}\frac{t-\left\{ t\right\} }{\sqrt{2t-1}}dt$$ where $0\leq\left\{ t\right\} <1$ is the fractional part of $t$. So $$\frac{2}{3}N\sqrt{2N-1}-\frac{1}{3}\sqrt{2N-1}+\frac{2}{3}<\sum_{k=1}^{N}\sqrt{2k-1}<\frac{2}{3}N\sqrt{2N-1}+\frac{2}{3}\sqrt{2N-1}-\frac{1}{3}.$$