Is it true that $ \sum_{i=1}^m \frac{1}{\sqrt{i}} = O \left( \sqrt{ m-1 } \right) $?

Question

Is it true that?: $$ \sum_{i=n}^m \frac{1}{\sqrt{i}} = O \left( \sqrt{ \frac{m-n}{n}} \right) $$ In special case if we have $n = 1$, is it true that?: $$ \sum_{i=1}^m \frac{1}{\sqrt{i}} = O \left( \sqrt{ m-1 } \right) $$

Answer 1

Fix $n \in \mathbb{N}-\left\{0\right\}$. One may use Stolz–Cesàro theorem: $$\lim_{m \to \infty} \frac{\sum \limits_{k=n}^{m} \frac{1}{\sqrt k}}{2\sqrt{ m-n}} = \lim_{m \rightarrow \infty}\frac{\frac{1}{\sqrt m}}{2(\sqrt{ m-n} - \sqrt{ m-n-1})} = \frac{1}{2} \lim_{n \rightarrow \infty} \frac{\sqrt{ m-n} + \sqrt{ m-n-1}}{\sqrt m} = 1,$$ giving, as $m \to \infty$,

$$ \sum_{i=n}^m \frac{1}{\sqrt{i}} = O \left( \sqrt{ \frac{m-n}{n}} \right)=O \left( \sqrt{ m} \right) $$

as announced.

Answer 2

The other answer addresses your general question (with a big hammer). A more elementary method for your other asymptotic estimate is ...

Observe that for $i \geq 1$, $\frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i+1}} < \frac{1}{i}$, so the error in replacing $\sum_{i=1}^m i^{-1/2}$ with the integral of the same thing is at most $\ln m$. But, $\int_1^m i^{-1/2} \,\mathrm{d}i = 2(\sqrt{m}-1)$ and $\lim_{m \rightarrow \infty} \frac{\ln m}{\sqrt{m}} = 0$, so the error is asymptotically zero. I.e., $$ \sum_{i=1}^m i^{-1/2} \in O(\sqrt{m}) = O(\sqrt{m-1}) \text{.} $$