Which convergence test should be used for this series? The series is, $$\sum \limits_{n = 1}^\infty \frac{ \ln(1 + \frac{5}{n})}{\sqrt [3] {n+1}}.$$
Using the asymptotic equivalence concept I came up with,
$$\frac{ \ln(1 + \frac{5}{n})}{\sqrt [3] {n+1}} \sim \frac{ \ln(\frac{1}{n})}{\sqrt [3]{n}} (n \to \infty)$$
But now, I can't figure out what test to use. I have tried the comparation test and the limit comparation test and I couldn't do it. Can you give me a hint? Thanks.
 A: There is an error in your estimate.  It should be $\frac{ \frac{1}{n}}{\sqrt [3]{n}}$ rather than $\frac{ \ln(\frac{1}{n})}{\sqrt [3]{n}}$. Therefore the series converges by comparison with a $p$-series.
A: You used wrongly the asymptotic equivalence: recall that
$$\log(1+x)\sim_0 x$$
so we have
$$\frac{ \ln(1 + \frac{5}{n})}{\sqrt [3] {n+1}}\sim_\infty \frac{5}{n\sqrt [3] {n}}=\frac{5}{n^{4/3}}$$
and the Riemann series $\displaystyle \sum_{n\ge1}\frac{5}{n^{4/3}}$ is convergent so by asymptotic comparison the given series is convergent.
A: Note that
$$
0<\ln \Big(1+\frac{5}{n}\Big)<\frac{5}{n},
$$
and 
$$
\frac{1}{\sqrt[3]{n+1}}<\frac{1}{\sqrt[3]{n}},
$$
and hence
$$
0<\frac{\ln \!\Big(\!1+\frac{5}{n}\!\Big)}{\sqrt[3]{n+1}}<\frac{\frac{5}{n}}{\sqrt[3]{n}}=\frac{5}{n^{4/3}}.
$$
Comparison test and the fact that $\sum_{n=1}^\infty\frac{1}{n^a}$ converges iff $a>1$, provides convergence of
$$
\sum_{n=1}^\infty \frac{\ln \!\Big(\!1+\frac{5}{n}\!\Big)}{\sqrt[3]{n+1}}.
$$ 
A: Hint: $\ln(1+x)\sim x$ when $x\to0$. Then compare with the harmonic series.
