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I am trying to solve this using an integral test, but I am unsure whether or not this is correct.

Let $f:[2,\infty)\to\mathbb{R}$ be defined by $f(t)=\frac{\ln{(t)}}{t} >0\ \forall t\geq2$.

Now $f'(t)=\frac{1-\ln{(t)}}{t^2} < 0 \ \forall t> e$.

$\implies f(t)$ is positive and decreasing $\forall t>e$.

Let $N_e \in \mathbb{N}$ be s.t. $N_e>e$.

We may now use the integral test to determine if $\sum_{n=N_e}^\infty \frac{\ln{(n)}}{n}$ converges or diverges.

$$\int\limits_{N_e}^\infty \frac{\ln{(n)}}{n}dn = \lim_{b\to\infty}\int\limits_{N_e}^b \frac{\ln{(n)}}{n}dn $$ Let $u=\ln{(n)} \implies du=\frac{1}{n}dn$

$$\implies \lim_{b\to\infty}\int\limits_{N_e}^b \frac{\ln{(n)}}{n}dn =\lim_{b\to \infty} \int\limits_{\ln{(N_e)}}^{\ln{(b)}}udu \\ = \lim_{b\to \infty}[\underbrace{\frac{1}{2}\ln^2{(b)}}_\text{Diverges}-\underbrace{\frac{1}{2}\ln^2{(N_e)}}_\text{Converges}]$$ Thus $\sum_{n=N_e}^\infty \frac{\ln{(n)}}{n}$ diverges $\implies \sum_{n=1}^\infty \frac{\ln{(n)}}{n}$ diverges.

user860374
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1 Answers1

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This looks great. You checked that $f(t)$ was monotone decreasing, then showed that the improper integral diverged.

However, you could've gotten to your result a lot faster with the comparison test. For all $n > e$, we have:

$$\frac{\ln(n)}{n} > \frac{1}{n}$$

Since $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ diverges, then...

Kaj Hansen
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  • Thank you! :). The only reason I used the integral test, was because this came out of my set of notes asking "Can the integral test be used to test for convergence for the following series... and if yes, test it using the integral test" – user860374 Oct 21 '14 at 23:17
  • Glad I could help! It comes down to personal preference in a lot of cases. – Kaj Hansen Oct 21 '14 at 23:23