Asymptotics for $p$-series with $p=1/2$

Question

Reading solutions to a practice exam, and I come across this: $$ O\left(\sum_{d \leq \sqrt{x}} {1 \over \sqrt{d}}\right) = O\left(x^{1/4}\right). $$ There are $O(\sqrt{x})$ terms in the sum, which are bounded between $x^{-1/4}$ and $1$. It looks like using the number of terms and the lower bound would give this estimate, but I don't see how that's valid.

Answer 1

By comparison to integrals, $$ 2\sqrt{n+1}-2=\int_1^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\le\sum_{k=1}^n\frac1{\sqrt{k}}\le\int_0^n\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{n} $$ So $$ \sum_{k=1}^n\frac1{\sqrt{k}}=O(\sqrt{n}) $$ plug in $n=\sqrt{d}$.