Here's a small introduction on what I am doing skip to the %%%%%%%%%% if you just want the question.
Definition: We say that $f(z) << g(z)$ if $|f(z)| \leq |g(z)| \quad \forall z \in D$.
In the book I'm reading (Analytic Number Theory, Donald J. Newman, page 25), they proved the fact that
$$p(n) - q(n) << M\frac{((1-r)^{5/2}}{r^n}\exp\left(\frac{\pi^2}{6}\frac{1}{1-r}\right).$$
With a clever choice of $r = 1- \frac{\pi}{\sqrt{6n}}$ they want to show that
$$p(n) = q(n) + O\left(n^{-5/4} e^{\pi\sqrt{2n/3}}\right).$$
It is clear for me that the $n^{-5/4}$ comes from the $(1-r)^{5/2}.$ It is also clear for me that $e^{\pi\sqrt{n/6}}$ comes from the $\exp\left(\frac{\pi^2}{6}\frac{1}{1-r}\right)$.
If I could show that $$\frac{1}{r^n} = \frac{1}{(1-\frac{\pi}{\sqrt{6n}})^n}$$ brings another $$e^{\pi\sqrt{n/6}}$$ in the $O$, I would be done with my understanding on this chapter.
I first thought that I could say $$\frac{1}{(1-\frac{\pi}{\sqrt{6n}})^n} = \frac{1}{((1-\frac{\pi}{\sqrt{6n}})^\sqrt{n})^\sqrt{n}} =(e^{\frac{\pi}{\sqrt{6}}})^{\sqrt{n}}=e^{\pi\sqrt{n/6}}.$$
But I did calculate the limit in one part of the equation and letting a $n$ untouched.
I could also try to show that $$\lim\limits_{n\rightarrow \infty} \frac{e^{\frac{-\pi\sqrt{n}}{\sqrt{6}}}}{(1-\frac{\pi}{\sqrt{6n}})^n} = k \quad \quad k \in \mathcal{R}$$
Then I would have a clean argument for my question which is:
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Is there anyone who could write me a valid argument for the fact that $$\frac{1}{r^n} = \frac{1}{(1-\frac{\pi}{\sqrt{6n}})^n} = O(e^{\pi\sqrt{n/6}}).$$
(This is my first time manipulating big $O$ notation.)
Start by writing $$\tag{1} \frac{1}{(1-\frac{\pi}{\sqrt{6n}})^n}=\exp\left[-n\log\left(1-\frac{\pi}{\sqrt{6n}}\right)\right] $$ Now, recall the power series for $\log(1-x)$: $$ \log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n},\qquad\lvert x\rvert<1. $$ In particular, $\log(1-x)=-x+O(x^2)$, so that $$ \log\left(1-\frac{\pi}{\sqrt{6n}}\right)=-\frac{\pi}{\sqrt{6n}}+O\left(\frac{1}{n}\right). $$ Plugging this in to (1), we find $$ \frac{1}{(1-\frac{\pi}{\sqrt{6n}})^n}=\exp\left[\pi\sqrt{\frac{n}{6}}+O(1)\right]=O(e^{\pi\sqrt{n/6}}), $$ as desired. (This last step holds because $e^{O(1)}$ is, itself, $O(1)$ -- that is, bounded above by a constant.)