Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have


*

*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$

*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim  \dfrac{e}{2n}$


any help would be appreciated
 A: Using Binomial Exapnsion:

$$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}\\
=(n^2-1)^{-1/2}-(n^2+1)^{-1/2}\\
=h[(1-h^2)^{-1/2}-(1+h^2)^{-1/2}]\quad h:=n^{-1}\\
=h[(1+h^2/2+...)-(1-h^2/2+...)]\\
=h[h^2+...]\\
=O(h^3)=O(n^{-3})$$

And using the $e$ and $\ln$-Series:

$$e-\left(1+\frac1n\right)^n\\
=e-e^{\frac{\ln(1+1/n)}{1/n}}\\
=e-e^{1-1/(2n)+...}=e[1-e^{-1/(2n)+...}]\\
=e[1-(1+(-1/2n)+...)]\\
\sim e/(2n)$$

A: Hint: $\left(\dfrac{1}{n^2-1}\right)^{1/2} = \left(n^2-1\right)^{-1/2} = \dfrac{1}{n}\left(1-\frac{1}{n^2}\right)^{-1/2} = \dfrac{1}{n} + \mathcal{O}\left(\frac{1}{n^3}\right)$
A: $$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=$$ 
$$\dfrac{\sqrt{n^2+1}-\sqrt{n^2-1}}{\sqrt{n^2-1}\sqrt{n^2+1}}=$$
$$\dfrac{(\sqrt{n^2+1}-\sqrt{n^2-1})(\sqrt{n^2+1}+\sqrt{n^2-1})}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{n^2+1-n^2+1}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{2}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{2}{\sqrt{n^4-1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=O\left(\frac{1}{n^3}\right).$$
A: For the first one:
$$\begin{align}
\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}} &= \frac{1}{n}\left(\frac{1}{\sqrt{1-\frac{1}{n^2}}}-\frac{1}{\sqrt{1+\frac{1}{n^2}}}\right) \\
&= \frac{1}{n}\left(\frac{1}{1-\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)}-\frac{1}{1+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)}\right) \\
&=\frac{1}{n}\left(\left(1+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right)-\left(1-\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right)\right) \\
&=\frac{1}{n}\left(\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right) \\
&=\frac{1}{n}\left(\frac{1}{n^2}+o\!\left(\frac{1}{n^2}\right)\right) \\
&=\frac{1}{n^3}+o\!\left(\frac{1}{n^3}\right)
\end{align}$$
using the Taylor expansions:


*

*$\sqrt{1+x} \operatorname*{=}_{x\to0} 1+\frac{x}{2} +o(x)$

*$\frac{1}{1+x} \operatorname*{=}_{x\to0} 1-x+x^2-\cdots+x^k +o(x^k)$

*$\frac{1}{1-x} \operatorname*{=}_{x\to0} 1+x+x^2+\cdots+x^k +o(x^k)$



For the second:
$$\begin{align}
e - \left(1+\frac{1}{n}\right)^n &= e - e^{n\ln\left(1+\frac{1}{n}\right)}
= e - e^{n\left(\frac{1}{n}-\frac{1}{2n^2} + o\!\left(\frac{1}{n^2}\right)\right)} \\
&= e^1 - e^{1-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)} \\
&= e\cdot\left( 1 - e^{-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)}\right) \\
&= e\cdot\left( 1 - \left( 1-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)\right)\right) \\
&= e\cdot\left( \frac{1}{2n} + o\!\left(\frac{1}{n}\right)\right) \\
\end{align}$$
using this time the Taylor expansions:


*

*$\ln(1+x) \operatorname*{=}_{x\to0} x - \frac{x^2}{2} +o(x^2)$

*$e^x \operatorname*{=}_{x\to0} 1+x+o(x)$

