Test whether $\sum_{n=1}^{\infty}\frac{\ln{n}}{n}$ converges or diverges

Question

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.

Answer 1

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...