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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.

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    $\begingroup$ How about the series $\frac{1}{n \ln n}$? Since the harmonic sum is asymptotic to $\ln n$ (more precisely, $\ln n + \gamma$, but this doesn't matter), the given series diverges iff $\sum \frac{1}{n \ln n}$ diverges. $\endgroup$ – Srivatsan Oct 9 '11 at 19:38
  • $\begingroup$ That seems appropriate, however the book has not covered this fact (the asymptotic relationship of the harmonic sum and $\ln n$). $\endgroup$ – process91 Oct 9 '11 at 19:43
  • $\begingroup$ Are you familiar with the integral test? That will tell you that $\sum_2\infty \frac{1}{n\ln n}$ diverges. It is not hard to show that $1+1/2+\cdots+1/n$ is $\lt c\ln n$ for suitable $c$, which will then give the result you want. For the integral test part, you can explicitly evaluate $\int_2^M \frac{1}{x\ln x}\,dx$. $\endgroup$ – André Nicolas Oct 9 '11 at 19:43
  • $\begingroup$ Alright, seems like that could be what the author intended, I will attempt to show that $1+1/2+\dots+1/n<c\log n$ $\endgroup$ – process91 Oct 9 '11 at 19:46
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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.

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  • $\begingroup$ Where can I find a proof of this theorem? Nevermind, found it - Cauchy condensation test. $\endgroup$ – process91 Oct 9 '11 at 20:15
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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.

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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.

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  • $\begingroup$ I like this method very much. I ended up using the Cauchy condensation test, but this is a very nice proof as well - thanks! $\endgroup$ – process91 Oct 10 '11 at 2:04
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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.

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