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

Question

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

Answer 1

Hint: $\left(\dfrac{1}{n^2-1}\right)^{1/2} = \left(n^2-1\right)^{-1/2} = \dfrac{1}{n}\left(1-\frac{1}{n^2}\right)^{-1/2} = \dfrac{1}{n} + \mathcal{O}\left(\frac{1}{n^3}\right)$

Answer 2

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:

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

Answer 3

$$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=$$ $$\dfrac{\sqrt{n^2+1}-\sqrt{n^2-1}}{\sqrt{n^2-1}\sqrt{n^2+1}}=$$ $$\dfrac{(\sqrt{n^2+1}-\sqrt{n^2-1})(\sqrt{n^2+1}+\sqrt{n^2-1})}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$ $$\dfrac{n^2+1-n^2+1}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$ $$\dfrac{2}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$ $$\dfrac{2}{\sqrt{n^4-1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=O\left(\frac{1}{n^3}\right).$$

Answer 4

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:

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

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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:

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