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Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$

I need to show that there exists $N \in \mathbb{N}: \forall n \geq N : \left|(1-\frac{1}{n^2})^n-1\right| \lt \epsilon$ $,\:\:\forall \epsilon \gt 0.$

Since $(1-\frac{1}{n^2})^n \leq 1$ $\forall n \in \mathbb{N}_+ \Rightarrow \left|(1-\frac{1}{n^2})^n-1\right| = 1 - (1-\frac{1}{n^2})^n$

Therefore I need to solve $1 - \left(1-\frac{1}{n^2}\right)^n < \epsilon$ for $n$.

Unfortunately I cannot solve this explicitly for n. The best I can do is:

$$\log(1-\epsilon) \lt n \log\left(1-\frac{1}{n^2}\right)$$

Is there a way to solve this for $n$ explicitly? How can I prove this if this is if it is not solvable for $n$?

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Do you know that $(1 + 1/n)^n \rightarrow e$? – PhoemueX May 18 '14 at 12:04
up vote 2 down vote accepted

Consider the expansion of $(1-\frac{1}{n^2})^n.$

$$\begin{align}\left(1-\frac{1}{n^2}\right)^n &= 1 - n\dfrac{1}{n^2}+\dfrac{n(n-1)}{2}\dfrac{1}{n^4} - \cdots \\ &= 1-\dfrac{1}{n}+\dfrac{n-1}{2n^3}+\cdots\\&\geq1-\dfrac{1}{n}\end{align}$$

We can construct the inequality $$1-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n \leq 1$$

Subtract $1$ and then change the order of inequality

$$-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n -1 \leq 0$$

$$0 \leq 1- \left(1-\frac{1}{n^2}\right)^n \leq \dfrac{1}{n}$$

After this you can just choose $N = \left\lceil\dfrac{1}{\epsilon}\right\rceil.$

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I am not sure that the expansion you use and the way you use it are enough to prove the inequality $(1-1/{n^2})^n\geqslant1-1/n$ (which holds nevertheless). You might want to make this point more precise. – Did May 18 '14 at 14:01
@Did I will amend my post to make the inequality more clear. On a related note, would it be an issue to expand around $n = \infty$? – user147887 May 18 '14 at 14:15

Hint: if $x\geqslant 0$ and $n$ is an integer, then $(1-x)^n\geqslant 1-nx$.

This can be seen definigin $f(x)=(1-x)^n+nx$ and showing that the derivative is non-negative.

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Hint: \begin{equation} \lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{\Large n}=\lim_{n\to\infty}\left[\left(1-\frac{1}{n^2}\right)^{\Large n^2}\right]^{\Large\frac{1}{n}}, \end{equation} where $\displaystyle\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{\Large n^2}=\frac {1}{e}$.

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Thanks Mr. @Eleven-Eleven. No one but you notice it, luckily I didn't get downvote \(^◡^)/ – Anastasiya-Romanova 秀 May 18 '14 at 13:05
Did you downvote my answer Mr. @Eleven-Eleven? – Anastasiya-Romanova 秀 May 18 '14 at 13:18
I'm not the downvoter, but I'm not convinced. Why are you allowed to take the limit of the expression in square brackets first, before taking the limit including the power $1/n$? – SpamIAm May 18 '14 at 14:13
@SpamIAm $$ \lim_{n\to\infty}\left[\left(1-\frac{1}{n^2}\right)^{ n^2}\right]^{\frac{1}{n}}=\left[\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n‌​^2}\right]^{\lim_{n\to\infty}\frac{1}{n}} $$ – Anastasiya-Romanova 秀 May 18 '14 at 15:02
Heuristically, $\left(\,1 - {1 \over n^{2}}\,\right)^{n} = \exp\left(\,n\ln\left(\,1 - {1 \over n^{2}}\,\right)\,\right) \sim\exp\left(\,-\,{1 \over n}\,\right)$ when $n \gg 1$. – Felix Marin Sep 4 '14 at 6:35

Use the taylor series bound $|\log(1-x)| \leq x + C|x|^2$ to show that $$\limsup_{n \to \infty} \left|n \log\left(1-\frac{1}{n^2} \right) \right| \leq \limsup_{n \to \infty} n \left(\frac{1}{n^2} + \frac{C}{n^4} \right) = 0,$$ hence $$\exp \left(n \log\left(1-\frac{1}{n^2} \right) \right)=\left(1-\frac{1}{n^2}\right)^n \to \exp(0) = 1.$$

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