Let $f(x) := \sum_{k = 1}^{N}1/|x + k|$ for $x \in [0, N]$. Why is $f(x) \geq C\log N$ for all $x \in [0, N]$ where $C$ is an absolute constant.
My work is: Since $x \in [0, N]$, we can remove the absolute value sign. Since each $1/(x + k)$ is montonically decreasing, $1/(x + k) \geq 1/(N + k)$. Then $$f(x) \geq \sum_{k = 1}^{N}1/(N + k) = \frac{1}{N + 1} + \cdots + \frac{1}{2N}$$ but isn't the right hand side above roughly $\log 2$? I don't know how to show the lower bound of $\log N$.