Showing that a series is convergent So I have the following question:

Show that the following series are convergent:

a) $\displaystyle\sum_{n=2}^{\infty} \frac{\log_n(n!)}{n^3}$
b) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(3n-2)^{n+\frac{1}{2}}}$
I'm not really sure how to approach this. For the first one, I guess I could use the ratio test since I know that the $n!$ can be replaced with $n(n-1)(n-2)...$ and using the ratio test will just cancel terms and I can take a limit.
For the 2nd sum, could I use the root test?
Any hint to just start off would be awesome.
 A: For a), note
$$\begin{equation}\begin{aligned}
\log_n(n!) & = \sum_{i=2}^{n}\log_n(i) \\
& \le \sum_{i=2}^{n}\log_n(n) \\
& = \sum_{i=2}^{n}1 \\
& = n - 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus,
$$\frac{\log_n(n!)}{n^3} \le \frac{n-1}{n^3} \lt \frac{1}{n^2} \tag{2}\label{eq2A}$$
For b), for $n \ge 2$, you have that
$$\frac{1}{(3n-2)^{n+\frac{1}{2}}} \lt \frac{1}{n^2} \tag{3}\label{eq3A}$$
Thus, in both cases, you can use the direct comparison test with the sums being $\gt 0$ as all the summation terms are positive and less than $\sum_{n=1}^{\infty} \frac{1}{n^2}$.
A: For a), the ratio test is actually inconclusive.
Instead, we have the simple bound of $n!\le n^n, as is clear from:
\begin{align}n!&=1\times2\times3\times\dots\times n\\n^n&=\underbrace{n\times n\times n\times\cdots\times n}_n\end{align}
Thus,
$$\frac{\log_n(n!)}{n^3}\le\frac n{n^3}=\frac1{n^2}$$
So convergence can be proven by a simple comparison.

For b), the root test is indeed applicable. Note that the exponent tends to 1 and that the whole denominator will tend to infinity.
