Proof that series diverges Prove that $\displaystyle\sum_{n=1}^\infty\frac{1}{n(1+1/2+\cdots+1/n)}$ diverges.
I think the only way to prove this is to find another series to compare using the comparison or limit tests. So far, I have been unable to find such a series.
 A: This answer is similar in spirit to Didier Piau's answer.
The following theorem is a very useful tool:
Suppose that $a_k > 0$ form a decreasing sequence of real numbers. Then
$$\sum_{k=1}^\infty a_k$$ converges if and only if $$\sum_{k=1}^\infty 2^k a_{2^k}$$ converges.
Applying this to the problem in hand we are reduced to investigating the convergence of
$$\sum_{k=1}^\infty \frac{1}{1 + 1/2 + \dots + 1/2^k}$$
But one easily sees that
$$1 + 1/2 + \dots + 1/2^k \le 1 + 2 \cdot 1/2 + 4 \cdot 1/4 + \dots + 2^{k-1}\cdot1/2^{k-1} + 1/2^k \le k + 1.$$
Because
$$\sum_{k=1}^\infty \frac{1}{k+1}$$
diverges, we are done.
A: Hints: 1. For every $k\geqslant0$, the sum of $\frac1n$ from $n=2^k$ to $n=2^{k+1}-1$ is at most $1$. 2. For every $k\geqslant0$ and every $n$ such that $2^k\leqslant n\leqslant 2^{k+1}-1$,  $1+\frac12+\cdots+\frac1n\leqslant k+1$ and $\frac1n\geqslant\frac1{2^k}$. 3. For every $k\geqslant0$, the terms from $n=2^k$ to $n=2^{k+1}-1$ of the series which interests you sum to at least $\frac1{k+1}$. 4. The harmonic series diverges.
A: Let $H_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n}$. $H_n$ is known as harmonic sum. Note that $H_{n+1} - H_n = \frac{1}{n+1}$. Now, consider $d_n = H_n - 1 - \ln(n)$. Then $d_{n+1} - d_n = \frac{1}{n+1} + \ln\left( 1 - \frac{1}{n+1} \right) < 0$,  $\forall n \ge 1$ which follows from $\ln(1-x) < -x$, for $0<x<1$. 
Now, since $d_{n+1} - d_n < 0$ and $d_1 = 0$, it follows that $H_n \le 1 + \ln(n)$, and thus 
$\sum_{n=1}^m \frac{1}{n H_n} \ge \sum_{n=1}^m \frac{1}{n( 1 + \ln(n))}$. Since the latter diverges by integral test, so does the original sum.
A: A possible approach thing to do is show this is greater than $$\int_{x=1}^{n+1} \frac{1}{x \log_e(x)}dx = \log_e(\log_e(n+1))$$ or some multiple of it, and show that the later diverges as $n$ increases.
