Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent

Question

How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?

I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{1}{(n+1)+\frac{(n+1)}{2}+...+1}\right)= 1$$ Then I thought using abel or dirichlet test. But I couldn't solve it.

Answer 1

Your sum is

$$\sum_{n=1}^\infty \frac{1}{n H_n},$$

where $H_n = \sum_{k=1}^n \frac{1}{k}$ is the nth harmonic number.

Since $1/x > 1/(k+1)$ for $x \in [k,k+1]$,

$$\ln n = \int_1^n \frac{1}{x} \,dx \geq \sum_{n=2}^{n+1} \frac{1}{k} = H_{n+1} - 1 > H_n - 1.$$

Now use the comparison test.

Answer 2

First, We need a lemma.

Lemma:

For $n>1$, $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1} < n$$

This could be checked easily with induction.

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To prove the question we use Cauchy Condensation Test, and see that its equal to show $$\sum_{k=1}^{\infty}1+\frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{2^k}$$ is divergent.

It can be checked easily that:

$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{k+1}-1}}$$

Now we apply our lemma for $k\geq 2$ and see:

$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}+...+\frac{1}{2^{k+1}-1}} > k+1$$

hence:

$$\sum_{k=1}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}} = \frac{1}{1+\frac{1}{2}} + \sum_{k=2}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}} > \sum_{k=3}^{\infty}{k}$$

Since the left hand side of inequality is divergent, the other side is so. Finally by Cauchy Condensation Test we see that the we have proved the divergence.

Answer 3

First note that $$ \sum\limits_{k = 1}^n {\frac{1}{k}} \le 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t} } } = 1 + \int_1^n {\frac{{dt}}{t}} = 1 + \log n, $$ for all $n\geq 1$. Then for all $N\geq 3$, \begin{align*} & \sum\limits_{n = 1}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}} > \sum\limits_{n = 3}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}} \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + 1)}}} \\ & \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + \log n)}}} = \frac{1}{2}\sum\limits_{n = 3}^N {\frac{1}{{n\log n}}} \\ & \ge \frac{1}{2}\sum\limits_{n = 3}^N {\int_{n}^{n+1} {\frac{{dt}}{{t\log t}}} } = \frac{1}{2}\int_3^{N+1} {\frac{{dt}}{{t\log t}}} \\ & = \frac{{\log \log( N+1) - \log \log 3}}{2} . \end{align*} This shows that the series diverges.