Convergence of two series, $\sum_{i=1}^\infty \frac{\sqrt{n+1}-\sqrt{n-1}}{n}$ and $\sum_{i=1}^\infty \frac{\ln(n!)}{n^a}$ I need help with $2$ series:

$$\sum_{i=1}^\infty \frac{\sqrt{n+1}-\sqrt{n-1}}{n} \tag{1}$$

I've gotten it into the form
$$\sum_{i=1}^\infty \frac{2}{n(\sqrt{n+1}+\sqrt{n-1})}$$
and I think that by the comparison test it converges just don't know what to compare it to.
Natural choice would be $$\sum_{i=1}^\infty \frac{{2}}{n(n+1)}$$ but it goes to zero faster so I don't know what to choose to compare it to or how to prove it converges. 

$$\sum_{i=1}^\infty \frac{\ln(n!)}{n^a} \tag{2}$$ for $a\in \mathbb{R}$.

I don't even know how to start here.
Thank you for your help.
 A: 1) The denominator of the sequence behaves like
$$ \frac 1{n \sqrt n}\,,$$
so you could use this for the comparison test.
2) One possibility is to apply Stirling's formula
$$ n! \sim \left(\frac ne\right)^n \sqrt{2\pi n} $$
which would allow you to simplify the logarithm.
A: It is easy to show that
$$\frac2{n(\sqrt{n+1}+\sqrt{n-1})}<\frac2{n\sqrt{n+1}}<\frac2{n\sqrt n}$$
Since subtracting things out of the denominator makes fractions bigger if everything is positive (think $1/0=+\infty$)
We may now apply direct comparison with the p-series to see that
$$\sum_{i=1}^\infty\frac2{n(\sqrt{n+1}+\sqrt{n-1})}<\sum_{i=1}^\infty\frac2{n\sqrt n}<\infty$$

The second one is not so hard if you notice that
$$\begin{align}\ln(n!)=\ln(1\times2\times\dots\times n)&=\sum_{k=1}^n\ln(k)\\&<\sum_{k=1}^n\ln(n)\\&=n\ln(n)\end{align}$$
We thus conclude from the direct comparison, Cauchy condensation test and geometric series that it converges if the following converges:
$$\sum_{k=1}^\infty\frac{\ln(k!)}{k^a}<\sum_{k=1}^\infty\frac{\ln(k)}{k^{a-1}}<\sum_{k=0}^\infty\frac{2^k\ln(2^k)}{(2^k)^{a-1}}=\sum_{k=0}^\infty\frac{k\ln(2)}{2^{k(a-2)}}<\infty$$
Showing that it diverges for $a\le2$ may likewise be done by rearranging the Riemann sums such that we get lower bounds instead of upper bounds.
A: As usual, straight asymptotics yields a quick solution.


*

*$\displaystyle \frac{2}{n({\sqrt{n+1}}+\sqrt{n-1})}\sim\frac{1}{n\sqrt n}=\frac{1}{n^{3/2}}$ hence convergence of the series.

*With the well-known estimate $\ln(n!)\sim n\ln n$, $\displaystyle \frac{{\ln(n!)}}{n^a}\sim \frac{\ln n}{n^{a-1}} $
As a result, the series converges if and only if $a-1>1$ ie $a>2$.

$\sim$ denotes asymptotic equivalence, as defined there.
Stirling's estimate is referenced there.
