testing if the infinite series are converges or diverges here are two problems on testing if the infinite series are conv. or div.

*

*$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$


*$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$
the first one I couldn't do :(
but the second one here is what I did using the ratio test I got that
$$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n)+\cdots+ \ln(2)}{\ln(n+1)+\cdots +\ln(2)}<1$$
which I thought would be effective but it isn't :(

In the first one, you have
\begin{align}
& \frac{n^{13/6} + \text{comparatively negligible things}}{n^3 + \text{comparatively negligible things}} \\[12pt]
= {} & \frac 1 {n^{5/6} + \text{comparatively negligible terms}} \ge \frac {\text{some constant}} {n^{5/6}} \\[6pt]
& \text{(and “constant” means not changing as $n$ changes)}
\end{align}
Here I would work on trying to prove that the $n$ term of the series is greater than or equal to some constant multiple of $n^{-5/6}.$
\begin{align}
& \frac{7\sqrt[6]{n^{13}}+2n}{\sqrt[3]{27n^9-10n+16}} = \frac{7 + \frac 2 {n^{7/6}}}{\sqrt[3]{27n^{5/2} + \frac {10} {n^{11/2}} + \frac{16}{n^{13/2}}}} \\[12pt]
\ge {} & \frac 7 {\sqrt[3]{27n^{5/2} + 10 n^{5/2} + 16n^{5/2}}} = \frac 7 {n^{5/6} \cdot\sqrt[3]{27+10+16}}
\end{align}
And $\displaystyle \sum_{n=1}^\infty \frac 1 {n^{5/6}} =  +\infty.$
