Q: Showing that the series $\sum \log{n}/n^2$ converges. I aim to show that the series $$\sum \frac{\log{n}}{n^2}$$ converges.
I know that the inequality $log(n) < \sqrt{n}$ holds for large $n$. So this give us one way to prove the convergence of $\sum \log{n}/n^2$ (by comparison). But to obtain this inequality we have to define log rigorously and put more advanced stuff in the game. Can we demonstrate without using this kind of properties, without continuity, derivatives and integrals? You can use simple properties of log (of the sum or product), the limit $\log{n}/n\to 0$, etc.

A: You can observe that, since $\log{n} \leq n $ for all $ n $,
$$\log{n} = \log{\sqrt{n}^2} = 2\log{\sqrt{n}} \leq 2\sqrt{n}$$
Asimptotically, the $ 2 $ changes nothing

A: You could use the integral test, since $\int_1^\infty \frac{\ln x \text{d}x}{x^2}=\int_0^\infty u\text{e}^{-u} \text{d}u=\left[\left(-1-u\right)\text{e}^{-u}\right]_0^\infty=1.$

A: You can use the Cantor criterion:

Let $\{ a_n\}_n$ be a positive decreasing sequence of real numbers.
$\sum_n a_n$ converges if and only if $\sum_n 2^n a_{2^n}$ converges.

Applying this criterion to your series, you need to show that
$$\sum_{n=1}^{\infty} 2^n \frac{\log 2^n}{4^n}$$
converges. But now
$$\sum_{n=1}^{\infty} 2^n \frac{\log 2^n}{4^n} = \log 2 \sum_{n=1}^{\infty}  \frac{n}{2^n}$$
is convergent by the root test.
