Why do we have $\sum_{m \le \sqrt x} \frac{1}{m} \gg \log x$ for $x\ge 1$?
I thought we could compare the sum with the integral and since $1/t$ is decreasing we have $\sum_{m\le \sqrt x}\frac{1}{m} \ll \int_1^\sqrt{x} \frac{1}{t}dt = \log {\sqrt{x}}$. What am I doing wrong here?