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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?

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  • $\begingroup$ Hint: $0 < H_m \operatorname*{\sim}_{m\to\infty}\ln m$. $\endgroup$ – Clement C. Jun 27 '14 at 10:52
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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.

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  • $\begingroup$ Thanks. For the $C$ one can take $\frac{1}{\log(2)} + 1$: An application of the integral test for finite sums gives $H_{m} \leq 1 + \log(m)$. With the $C$ above it can be easily verified that $\log(m) + 1 \leq C\log(m)$ when $m \geq 2$. The result follows since $H_{m-1} \leq H_{m}$ for all $m$. $\endgroup$ – ttb Jun 27 '14 at 21:46
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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

$$\sum_{n=1}^{\infty} \{ \log s_{n+1} - \log s_{n} \} = \lim_{n\to\infty} \log s_{n} = \infty.$$

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