Convergence or divergence of $\sum_{n=1}^{\infty} \ln({1+ n^{-4/3}})$

Question

How to check if this series converges or diverges? $$\sum_{n=1}^{\infty} \ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)$$
I've tried using comparison test so.
$$c_n=\ln\left(\frac{1}{n^{4/3}}\right) <\ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)<\ln\left(1+\frac{1}{n}\right)=b_n$$
By integral test $c_n$ and $a_n$ is divergent, so initial series should be divergent too. But using wolfram mathematica it shows that series is convergent. Any ideas?

Answer 1

Recall that for any $x\ge0 \implies \log(1+x)\le x$, therefore

$$0\le \ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)\le \frac{1}{n\sqrt[3]{n}}$$

and since $\sum \frac{1}{n\sqrt[3]{n}}$ converges by $p$ test also the given series converges by comparison test.