Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$

Question

How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$

using the following method :

note that : $(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^{2})\quad ( x\to 0) $

\begin{align} \frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\ &\text{since } \sqrt{n+1}\sim \sqrt{n} \text{ then } \lim_{n\to+\infty}\frac{\sqrt{n+1}}{n}=0 \\ &\sim \frac{(-1)^n}{n}\left(1+\frac{(-1)^n\sqrt{n}}{n} \right)^{-1}\\ &= \frac{(-1)^{n}}n\left(1+\mathcal{O}\left( \frac{(-1)^n\sqrt{n}}{n}\right) \right)\\ &= \frac{(-1)^n}{n}+\mathcal{O}\left( \dfrac{(-1)^{2n}\sqrt{n}}{n^2}\right) \\ &=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right) \end{align}

• AM i right ?

Answer 1

One may write, as $n \to \infty$, $$ \begin{align} \frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\ &= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\cdot\sqrt{1+\frac1n}\: \right)^{-1}\\ &= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\left(1+\frac1{2n}+\mathcal{O}\left(\dfrac{1}{n^2}\right)\right)\: \right)^{-1}\\ &= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)^{-1}\\ &= \frac{(-1)^n}{n}\left(1-\frac{(-1)^n}{\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\: \right)\\ &=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right) \end{align} $$ as wanted.

Answer 2

I would initially ignore the $(-1)^n$ and do

$\begin{array}\\ \frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n} &=\frac{n-(n+(-1)^n\sqrt{n+1})}{n(n+(-1)^n\sqrt{n+1})}\\ &=\frac{-(-1)^n\sqrt{n+1}}{n(n+(-1)^n\sqrt{n+1})}\\ &=\frac1{n^{3/2}}\frac{(-1)^{n+1}\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\ \text{so}\\ \big|\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}\big| &=\frac1{n^{3/2}}\frac{\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\ &=O(\frac1{n^{3/2}})\\ \end{array} $

since $\sqrt{1+1/n} \to 1$ and $\sqrt{1/n+1/n^2} \to 0$.