I am reading a paper which claims that the following series diverges: $\sum\limits_{n=2}^{\infty}\frac{1}{nH_{n-1}}$ where $H_{n}$ is the $n$'th harmonic number $\sum\limits_{m=1}^{n}\frac{1}{m}$.
I tried a comparison test: Since $H_{n} < n$ each denominator $nH_{n-1} < n^{2}$ but that only lets me bound this from below by the convergent series $\sum\limits_{n}\frac{1}{n^{2}}$.
How can I show that this diverges?
On
Integral test argument combined with the comparison $H_{n} \sim \log n$ is quite famous, so here I demonstrate another possible approach:
Proposition. Suppose $a_{n} > 0$ and $\sum a_{n} = \infty$. If we denote $s_{n} = a_{1} + \cdots + a_{n}$, then $$ \sum_{n=1}^{\infty} \frac{a_{n}}{s_{n}} = \infty. $$
Proof. It suffices to prove when $a_{n}/s_{n} \to 0$. Note that this assumption implies
$$ \lim_{n\to\infty} \frac{a_{n}/s_{n}}{\log s_{n+1} - \log s_{n}} = \lim_{n\to\infty} \frac{a_{n}/s_{n}}{\log(1 + a_{n}/s_{n})} = 1. $$
Therefore the claim follows by the limit comparison test with the series
$$\sum_{n=1}^{\infty} \{ \log s_{n+1} - \log s_{n} \} = \lim_{n\to\infty} \log s_{n} = \infty.$$
Use the estimate $H_{n-1}\le C\log(n)$ for a suitable constant $C$.
Then it suffices to show that $\sum^\infty_{n=2} \frac{1}{n\log(n)}$ is divergent, which follows by the integral comparison test:
$$\int_2^R \frac{dt}{t\log(t)}=\log(\log(R))+\text{const.}\rightarrow \infty$$
as $R\rightarrow\infty$, albeit the convergence is extremely slow.