Asymptotics in exponential function This question begins with the following limit:
$$
\lim_{n\to+\infty}\left(\frac{\pi}{2}-\arctan n\right)^n n!.
$$
By using Stirling's formula, after some calculations, I obtain the following, equivalent, limit:
$$
\lim_{n\to+\infty}e^{n\ln\left(\frac\pi2 -\arctan n\right)+n\ln n-n}\sqrt{2\pi n}.
$$
Now, the exponent of the exponential function is
$$
n\ln\left(\frac\pi2 -\arctan n\right)+n\ln n-n =
n\left(\ln\left(n\left(\frac\pi2-\arctan n\right)\right)-1\right).
$$
Now,
$$
\lim_{n\to+\infty} n\left(\frac\pi2-\arctan n\right)=1,
$$
so
$$
n\ln\left(\frac\pi2 -\arctan n\right)+n\ln n-n\sim-n
$$
(where the symbol $\sim$ means asymptotically equivalent).
But I know that if $f(x)\sim g(x)$ it is not necessarily true that $e^{f(x)}\sim e^{g(x)}$, so I am not allowed to write that the original limit is equal to
$$
\lim_{n\to+\infty}e^{-n}\sqrt{2\pi n}=0.
$$
So what could I do now?
[EDIT]
After some discussion with @Virtuoz in comments, I found the following solution.
It is true that
$$
\arctan x + \arctan\frac{1}{x} = \frac\pi2,
$$
so the original limit becomes:
$$
\lim_{n\to+\infty}\left(\arctan\frac{1}{n}\right)^n n!.
$$
Now, it is also true that
$$
\arctan\frac{1}{n}\sim\frac{1}{n},
$$
but we have to prove that
$$
\left(\arctan\frac{1}{n}\right)^n\sim\left(\frac{1}{n}\right)^n.
$$
This can be proven as follows: if $\arctan(1/n)=y$, then $n=1/\arctan y$ and $y\to0^+$, so
\begin{align*}
\left(\frac{\arctan\frac{1}{n}}{\frac{1}{n}}\right)^n &=
\left(\frac{y}{\tan y}\right)^\frac{1}{\tan y}\\
&=\left(1+\frac{y}{\tan y}-1\right)^\frac{1}{\tan y}\\
&=\left( \left(1+\frac{y}{\tan y}-1\right)^\frac{1}{\frac{y}{\tan y}-1}\right)^{\left(\frac{y}{\tan y}-1\right)\frac{1}{\tan y}}.
\end{align*}
Now, the exponent
$$
\left(\frac{y}{\tan y}-1\right)\frac{1}{\tan y}=\frac{y\cos y -\sin y}{\sin^2 y}\cos^2 y\sim \frac{y-\frac12 y^3-y}{\sin^2 y}\cos^2 y\to 0,
$$
and this proves that
$$
\left(\frac{\arctan\frac{1}{n}}{\frac{1}{n}}\right)^n\to 0.
$$
So, finally,
$$
\lim_{n\to+\infty}\left(\arctan\frac{1}{n}\right)^n n! =
\lim_{n\to+\infty}\frac{n!}{n^n}=0.
$$
 A: Let $$g(n) = \log \left[\left(\arctan\frac{1}{n}\right)^n  n! \right]=n\log \arctan\frac{1}{n}+\log n!$$
Letting, $x=1/n$, using the Taylor expansion of $\arctan x$ at $x=0$:
$$ \arctan x = x + O(x^3)$$
we have
$$\log \arctan x =\log x +  \log \left(1 + O(x^2)\right)= \log(x) + O(x^2)$$
and
$$ g(n) = n \log n^{-1} +  O(n^{-1})+ n \log n -n + \Theta (\log n)=
          -n + \Theta (\log n)  \to - \infty$$
Hence
$$ \left(\arctan\frac{1}{n}\right)^n  n! = e^{g(n)} \to 0$$
A: You can use inequality
$$
\arctan x \ge \frac{x}{1 + \frac{2}{\pi} x}, \, x >0.
$$
This gives us the upper bound
$$
\left(\frac{\pi}{2}-\arctan n\right)^n n! \le \left( \frac{\pi}{2} - \frac{n}{1 + \frac{2}{\pi} n} \right)^n \cdot n! \sim \left( \frac{\pi^2}{2(\pi + 2n)} \right)^n \cdot \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \le
$$
$$
\le \left( \frac{\pi^2}{4n} \cdot \frac{n}{e} \right)^n \cdot \sqrt{2\pi n} \longrightarrow 0.
$$
The last expression tends to $0$ since
$$
\pi^2 < 4e.
$$
A: Just for you curiosity.
Starting from @leonbloy's answer to derive an asymptotics
$$\tan ^{-1}\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3 n^3}+\frac{1}{5
   n^5}+O\left(\frac{1}{n^7}\right)$$
$$\log \left(\tan ^{-1}\left(\frac{1}{n}\right)\right)=-\log (n)-\frac{1}{3 n^2}+\frac{13}{90   n^4}+O\left(\frac{1}{n^6}\right)$$
Using Stirling approximation
$$\log(n!)=n (\log (n)-1)+\frac{1}{2} (\log (n)+\log (2 \pi ))+\frac{1}{12
   n}-\frac{1}{360 n^3}+\frac{1}{1260
   n^5}+O\left(\frac{1}{n^7}\right)$$ All the above make
$$g(n)=-n+\frac{1}{2} (\log (n)+\log (2 \pi ))-\frac{1}{4 n}+\frac{17}{120
   n^3}+O\left(\frac{1}{n^5}\right)$$
$$e^{g(n)}=\sqrt{2 \pi n}\,\, e^{-n}\,\, \left(1-\frac{1}{4 n}+\frac{1}{32 n^2} +O\left(\frac{1}{n^3}\right)\right)$$ which is a quite good approximation. For example, using $n=10$ the "exact" value is $0.00035103$ while the above asymptotics leads to $0.00035099$
