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 :(
 A: Both are divergent. In the first one, the term is $\sim 7n^{13/6}/3n^3= \frac 73 n^{-5/6}$,
as one know $\sum n^{-p}$ is divergent when $p<1$. In the second one, the term is greater
than $\frac{1}{n\ln n}=\frac{1}{\ln n^n}$, you see $\int_2^\infty \frac{1}{x\ln x}dx$ is divergent so by the integral test $\sum  \frac{1}{n\ln n}$ is divergent, so $\sum \frac{1}{\ln n!}$ is more divergent.
A: 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 $\displaystyle \sum_{n=1}^\infty \frac 1 {n^{5/6}} =  +\infty.$
A: As @Yuval showed, the first one diverges. 
Since the integral test is not allowed, you can use direct comparison test: 
Comparing your series to the harmonic series $\sum_{k=1}^{\infty}\frac{1}{k}$, which diverges, gives us 
$\frac{7}{3}\frac{1}{k^{\frac{5}{6}}}>\frac{1}{k} \Rightarrow \sum_{k=1}^{\infty}\frac{7}{3}\frac{1}{k^{\frac{5}{6}}}>\sum_{k=1}^{\infty}\frac{1}{k}$. 
Now, by the direct comparison test, the series diverges.
Remarks: 
(1) I used the asymptotic order of your series, but it works similar with the original term. 
(2) Divergence of the harmonic series can be proven by direct comparison test as well. (https://en.wikipedia.org/wiki/Harmonic_series_(mathematics))
