How do I prove that sum of reciprocals of first $2^n$ natural nos is always greater than $\frac{n+1}{2}$

Question

How can prove this inequality \begin{equation} \sum_{r=1}^{2^n} \frac{1}{r}\geq \frac{n+1}{2} \end{equation} Without using induction... Want to have an insight on summation inequalities

Answer 1

For non-negative integers $j$ we have, $$\sum_{k=1}^{2^j} \frac{1}{2^j+k}\geq\sum_{k=1}^{2^j}\frac{1}{2^{j+1}}=\frac{2^j}{2^{j+1}}=\frac{1}{2}$$

Now, for $n\geq1$, $$\sum_{r=1}^{2^n}\frac{1}{r}=1+\sum_{j=0}^{n-1}\left( \sum_{k=1}^{2^j}\frac{1}{2^j+k}\right)\geq\left(1+\frac{n}{2}\right)=\frac{n+2}{2}$$

Equality holds for $n=1$.

This was the key observation of Nicole Oresme in his proof of divergence of the harmonic series $$\sum_{m=1}^{\infty}\frac{1}{m}$$ Check this link https://en.m.wikipedia.org/wiki/Nicole_Oresme

Probably this post will help Denis James.