How do I prove the inequality $H_{2^m}>1+\frac{m}{2}$?

Question

How do I prove that $\sum_{k=1}^{2^m}\frac{1}{k}>1+\frac{m}{2}$?

I have used it in an exercise but I want to know why it is like that.

Answer 1

Simple way: assuming $m\geq 2$, $$ \sum_{k=1}^{2^m}\frac{1}{k} = 1+\sum_{h=0}^{m-1}\color{red}{\sum_{k=2^h+1}^{2^{h+1}}\frac{1}{k}}>1+\sum_{h=0}^{m-1}\frac{2^h}{2^{h+1}}=1+\frac{m}{2} $$ since every red sum is made by $2^h$ terms, the smallest of them being $\frac{1}{2^{h+1}}$.
This is the key idea of Cauchy's condensation test.

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Now something more accurate. Since $\frac{1}{x}$ is decreasing, positive and convex on $\mathbb{R}^+$,
by the Hermite-Hadamard inequality we have: $$ \frac{1}{2}\cdot\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^m-1}+\frac{1}{2}\cdot\frac{1}{2^m}>\int_{1}^{2^m}\frac{dx}{x} = m\log(2) $$ hence: $$ H_{2^m}>\frac{1}{2}+m\log(2) $$ that is a stronger inequality, since $\log(2)\approx\frac{61}{88}>\frac{1}{2}.$