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

Question

I would like to show that : $$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$ by starting from the left side and get the right side

My Proof:

\begin{align*} (-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\left(\dfrac{1}{\sqrt{n+1}+\sqrt{n} }\right)\\ &=(-1)^{n}\left(\sqrt{n+1}+\sqrt{n} \right)^{-1}\\ &=(-1)^{n}\left(\sqrt{n}\left(1+\sqrt{1+\dfrac{1}{n}}\right) \right)^{-1}\\ &=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\dfrac{1}{n}\right)^{\dfrac{1}{2}} \right)^{-1}\\ &=\mbox{Note that :}\left(1+x\right)^{\alpha}=1+\mathcal{O}\left( x\right) \\ &=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\mathcal{O}\left(\dfrac{1}{n} \right)\right) \right)^{-1}\\ &=\dfrac{(-1)^{n}}{\sqrt{n}}\left(2+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\ &=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\ &=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\mathcal{O}\left(\dfrac{1}{n}\right)\right) \right)\\ &=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)\\ &=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)\\ \end{align*} $$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$

• Am i right ?

Answer 1

It is correct to me. In short, $$ \begin{align} (-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\frac1{\sqrt{n+1}+\sqrt{n}} \\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+\sqrt{1+1/n}} \\\\&=\frac{(-1)^n}{\sqrt{n}}\frac1{1+1+O(1/n)} \\\\&=\frac{(-1)^n}{2\sqrt{n}}\left(1-O(1/n) \right) \\\\&=\frac{(-1)^n}{2\sqrt{n}}+O\left(\frac1{n^{3/2}} \right). \end{align} $$

Answer 2

You van do it also by purely algebraic means: $$\begin{align} \frac{1}{\sqrt{n+1}+\sqrt{n}}&=\frac{1}{2\sqrt{n}}+\frac{\sqrt{n}-\sqrt{n+1}}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})}\\ &=\frac{1}{2\sqrt{n}}-\frac{1}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})^2}. \end{align}$$

Answer 3

Simply using series expansion will do:

\begin{align*} \sqrt{n+1}-\sqrt{n} &= \sqrt{n} \left( \sqrt{1+\frac{1}{n}}-1 \right) \\ &=\sqrt{n} \left( 1+\frac{1}{2n}-\frac{1}{8n^2}+\ldots-1 \right) \\ &= \frac{1}{2\sqrt{n}}-\frac{1}{8n^{3/2}}+\ldots \end{align*}