Can we prove partial sums $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ for all $N$ Can we prove $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ ?
I can figure out how to prove this where $N$ is a power of $2$ by adding the successive inequalities listed below (Cauchy condensation)
$$ \frac{1}{1} > 0 $$
$$ \frac{1}{2} \ge \frac{1}{2}$$
$$ \frac{1}{3} + \frac{1}{4} \ge \frac{1}{4} \cdot 2 = \frac{1}{2}$$
$$ \frac{1}{5} + \cdots + \frac{1}{8} \ge \frac{1}{8} \cdot 4 = \frac{1}{2}$$
$$ \cdots $$
$$ \cdots $$
I am having trouble figuring out how to do this for the partial sums when $N$ is not a power of $2$.
Click here to see a is a graph on desmos showing how the partial sums grow.
If there is a way to prove $\sum_{n=1}^{N} \frac{1}{n} \ge \log_e(N)$, that will also suffice.
Thanks!

Here is one way.
$$
\sum_{n=1}^N \frac{1}{n} = \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{n}
\geq \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{x}
= \int_{1}^{N+1}\frac{dx}{x} = \Big[\ln x\Big]_1^{N+1}
= \ln (N+1)
$$
(I write $\ln$ for the natural logarithm.)
