Evaluation of :$\lim_{n\to \infty} {(1+\frac{1}{\arctan(n)}})^{\arctan(n)}$

Question

Really i'm interesting to trigonometric form which is related to :$\lim_{n\to \infty} {(1+\frac{1}{n}})^{n}$ , Wolfram alpha show that this: $\lim_{n\to \infty} {(1+\frac{1}{\arctan(n)}})^{\arctan(n)}$ equal to $(\frac{2+\pi}{\pi})^{\frac{\pi}{2}}$ as shown here , I have tried the variable change $y=\arctan(n)$ to get the same form with $\lim_{n\to \infty} {(1+\frac{1}{n}})^{n}$ but I didn't succeeded , Then Is there any simple way to evaluate :$$\lim_{n\to \infty} {(1+\frac{1}{\arctan(n)}})^{\arctan(n)}$$

Answer 1

Consider $$A=\left(1+\frac{1}{\tan ^{-1}(n)}\right)^{\tan ^{-1}(n)}\implies \log(A)={\tan ^{-1}(n)}\log\left(1+\frac{1}{\tan ^{-1}(n)}\right)$$ Now, using Taylor expansion for large $n$ $$\tan ^{-1}(n)=\frac{\pi }{2}-\frac{1}{n}+O\left(\frac{1}{n^3}\right)$$ $$1+\frac{1}{\tan ^{-1}(n)}=\left(1+\frac{2}{\pi }\right)+\frac{4}{\pi ^2 n}+\frac{8}{\pi ^3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log\left(1+\frac{1}{\tan ^{-1}(n)}\right)=\log \left(1+\frac{2}{\pi }\right)+\frac{4}{\pi (2+\pi ) n}+O\left(\frac{1}{n^2}\right)$$ $$\log(A)=\frac{1}{2} \pi \log \left(1+\frac{2}{\pi }\right)+\frac{\frac{2}{2+\pi }-\log \left(1+\frac{2}{\pi }\right)}{n}+O\left(\frac{1}{n^2}\right)$$ So, $$\lim_{n\to \infty} \log(A)=\frac{1}{2} \pi \log \left(1+\frac{2}{\pi }\right)\implies \lim_{n\to \infty} A=\left(1+\frac{2}{\pi }\right)^{\pi /2}$$