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.

Answer 2

Since$$\lim_{x\to0}\frac{\log(1+x)}x=\log'1=1,$$you have$$\lim_{n\to\infty}\frac{\log\left(1+\frac1{n\sqrt[3]n}\right)}{\frac1{n\sqrt[3]n}}=1.$$So, since the series $\sum_{n=1}^\infty\frac1{n\sqrt[3]n}=\sum_{n=1}^\infty n^{-\frac43}$ converges, your series converges too.