Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$

Question

I would like to show that : $$(-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$

My proof:

Note that :
\begin{align*} e^{x}&=1+x+\mathcal{O}\left(x^{2}\right)\quad (x\to 0)\\ \tan(x)&=x+\mathcal{O}\left(x^{3} \right)\\ \tan(a+b)&=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\\ x^{a}&=e^{a\ln(x)}\quad (x>0) \end{align*} then:

• $-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)=\left(1-\tan^{-1}\left(\dfrac{1}{n}\right) \right)^{-1}$
• $\dfrac{1}{n} \underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow }0$
• $\left(\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^{3}} \right)\right)^{-1}=\left(\dfrac{1}{n}\right)^{-1}\left(1+\mathcal{O}\left( \dfrac{1}{n^{2}}\right) \right)^{-1}=n\left( 1-\mathcal{O}\left( \dfrac{1}{n^{2}}\right)\right)$

Thus :

\begin{align*} (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}&=(-1)^{n}e^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)\ln(n)} \\ &=(-1)^{n}\exp\left[-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)\ln(n)\right]\\ &=(-1)^{n}\exp\left[\left(1-\tan^{-1}\left(\dfrac{1}{n}\right) \right)^{-1}\ln(n)\right]\\ &=(-1)^{n}\exp\left[\left(1-\left(\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^{3}} \right)\right)^{-1} \right)^{-1}\ln(n)\right]\\ \end{align*}

I'm stuck

Answer 1

First, we note that $\tan\left(\frac{\pi}{4}+\frac1n\right)$ can be expanded as

$$\begin{align} \tan\left(\frac{\pi}{4}+\frac1n\right)&=\frac{1+\tan(1/n)}{1-\tan(1/n)}\\\\ &=1+\frac2n +\frac2{n^2}+O\left(\frac{1}{n^3}\right) \end{align}$$

Then,

$$\begin{align} e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{-\left(1+\frac2n +\frac2{n^2}+O\left(\frac{1}{n^3}\right) \right)\log(n)}\\\\ &=\frac1n\,e^{-\left(\frac{2\log(n)}n +\frac{2\log(n)}{n^2}+O\left(\frac{\log(n)}{n^3}\right)\right)}\\\\ &=\frac1n\,\left(1-2\frac{\log(n)}{n}-2\frac{\log(n)}{n^2}+O\left(\frac{\log^2(n)}{n^2}\right)\right)\\\\ &=\frac1n +O\left(\frac{\log(n)}{n^2}\right) \end{align}$$

Therefore, we find that

$$(-1)^n\,n^{-\tan\left(\frac{\pi}{4}+\frac1n\right)}=(-1)^n\frac1n +O\left(\frac{\log(n)}{n^2}\right)$$