Let $H_n=1+\frac12 +\frac13 +\cdots + \frac1n$. I want to show that the infinite sum $$\sum_{n=1}^{\infty} \dfrac{1}{H_n}$$ diverges.
My approach:
For $n,m\in\mathbb{N}$, is easy to verify that $H_n<H_m$ if $n<m$, thus $H_n<H_{2^n}$ for every $n\in\mathbb{N}$. But, \begin{align*} H_{2^n}&=1+\frac12 +\frac13 +\cdots + \dfrac{1}{2^n}\\ &<1+\left(\frac12+\frac12\right)+\left(\frac14+\frac14+\frac14+\frac14 \right)+\cdots+ \underbrace{\left(\dfrac{1}{2^{n-1}}+\cdots+\dfrac{1}{2^{n-1}}\right)}_{2^{n-1}\text{ times}}\\ &=1+1+\cdots +1=n \end{align*}
Then, $H_{2^n}\leqslant n$ for every $n$, so $$\frac1n\leqslant \dfrac{1}{H_n},$$ for every $n$ and since $$\sum_{n=1}^{\infty}\frac1n$$ diverges, the infinite sum $$\sum_{n=1}^{\infty}\dfrac{1}{H_{n}}$$ diverges too.
Am I right?
$\displaystyle\frac 1 {H_n} > \frac 1 n$, so divergence follows from the basic comparison test.