Stirling's Formula - Comparison Test Method The following question concerns the convergence of Stirling's Approximation for $n!$
I have $r_n = \frac{\sqrt{n}}{n!}(\frac{n}{e})^n$.
I have expressed $\log(\frac{r_{n+1}}{r_n})=(n+\frac12)\log(1+\frac1n)-1$,
Now the fun starts:
I am now told to evaluate the following limit:
$\lim_{x\to 0}\frac{(1+\frac{x}{2})\log(1+x)-x}{x^3} $, which I have found to be equal to $\frac{1}{12}$.
Assuming this is all fine, I am having trouble with the following question:
Using the simplified form of $\log(\frac{r_{n+1}}{r_n})$ and the limit found, show that:
$\sum_{n=1}^\infty \log(\frac{r_{n+1}}{r_n})$ converges, by using the Comparison Test.
Thoughts on this part:


*

*I believe this question has some relation to Taylor Series Expansion of $\log(1+x)$, since expanding $(n+\frac12)\log(1+\frac1n)-1$ gives $\frac{1}{12n^2}-\frac{1}{12n^3}+\frac{3}{40n^4}-\frac{1}{15n^5}+\ldots$ and expanding $\frac{(1+\frac{x}{2})\log(1+x)-x}{x^3}$ gives $\frac{1}{12}-\frac{x}{12}+\frac{3x^2}{40}-\frac{x^3}{15}+\ldots$ Pretty similar wouldn't you say?

*Another train of thought is that I am not sure how I can use the limit of something as $x$ tends to $0$ to evaluate some summation, even if they are related, since an infinite summation is the limit as $n$ tends to infinity of partial sums.


The next part, (which may just follow directly once I've figured out the previous part), is to then use this knowledge to prove that $\lim_{n\to \infty} r_n$ exists, althought obviously I've yet to try that.
This question has me a bit stumped and I'd really appreciate any tips or suggestions from the community!
Many Thanks
Benjie
 A: If you substitute $x=\frac1n$, the expression in the limit limit turns into
$$ \begin{align}\frac{(1+\frac x2)\log(1+x)-x}{x^3}&= \left(\Bigl(1+\frac1{2n}\Bigr)\log\Bigl(1+\frac1n\Bigr)-\frac1n\right)n^3\\& = \left(\Bigl(n+\frac1{2}\Bigr)\log\Bigl(1+\frac1n\Bigr)-1\right)n^2\\&=n^2\log\frac{r_{n+1}}{r_n}\end{align}.$$
As the $\lim_{x\to 0}$ of the original expression is $\frac1{12}$, so is the limit of the last expression as $n\to\infty$.
Especially, $\left|n^2\log\frac{r_{n+1}}{r_n}-\frac1{12}\right|<\frac{5}{12}$ for almost all $n$, hence 
$$ \left|\log\frac{r_{n+1}}{r_n}\right|<\frac1{2n^2}<\frac1{n(n+1)}=\frac1n-\frac1{n+1}$$
for almost all $n$. Thus the comparison test with the telescopic series $\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$ shows that $\sum_{n=1}^\infty\log\frac{r_{n+1}}{r_n}=\sum_{n=1}^\infty(\log r_{n+1}-\log {r_n})$ converges.
As the $n$th partial sum is just $\log r_{n+1}-\log r_1$ (again telescopic), the convergence of the series implies the convergence of $\log r_n$ to some limit $a$ and ultimately the convergence of $r_n$ to $e^a$.
A: You are totally correct, and you already more or less answered the question with that series in $1/n^k$.  Compare with 
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = \frac{\pi^2}{8}$$
Another thought is, isn't this just a telescoping series, so as long as the sequence converges, the series converges?
