Does $\sum{\ln n\over n^{4/3}}$ converge or diverge? Does $\displaystyle\sum{\ln n\over n^{4/3}}$ converge or diverge?
Which test should I use?
I tried the ratio test and root test but both of them are inconclusive.
I could try the comparison test but I don't know which function I have to compare with.
By the integral test you may compare your series with the following integral:
$$
\int_1^\infty \frac{\ln x}{x^{4/3}} dx=\left.\frac{x^{-4/3+1}}{-4/3+1}\ln x\right|_1^\infty-\frac1{-4/3+1}\int_1^\infty x^{-4/3+1}\frac1{x} dx=9&lt;\infty
$$ giving the convergence of your series
$$
\sum_1^\infty\frac{\ln n}{n^{4/3}}.
$$
An other way is to compare $\ln(.)$ with $\sqrt{.}$ in fact $$\forall x \in \mathbb{R}, \ln(x) \le \sqrt{x}$$

Then $$\forall n \in \mathbb{N}, \ln(n) \le \sqrt{n}$$

So for $n&gt; 1$ $$\left|\frac{\ln(n)}{n^{\frac{4}{3}}}\right|\le\left|\frac{\sqrt{n}}{n^{\frac{4}{3}}}\right|=\frac{1}{n^{\frac{6}{5}}}$$

And obviously $\frac{6}{5}&gt;1$ so the series is convergent.