3
$\begingroup$

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 $n$?

$\endgroup$
  • 1
    $\begingroup$ Do you know that $(1 + 1/n)^n \rightarrow e$? $\endgroup$ – PhoemueX May 18 '14 at 12:04
2
$\begingroup$

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 $N = \left\lceil\dfrac{1}{\epsilon}\right\rceil.$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Did May 18 '14 at 14:01
  • $\begingroup$ @Did I will amend my post to make the inequality more clear. On a related note, would it be an issue to expand around $n = \infty$? $\endgroup$ – user147887 May 18 '14 at 14:15
4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $\displaystyle\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{\Large n^2}=\frac {1}{e}$.

$\endgroup$
  • $\begingroup$ Thanks Mr. @Eleven-Eleven. No one but you notice it, luckily I didn't get downvote \(^◡^)/ $\endgroup$ – Anastasiya-Romanova 秀 May 18 '14 at 13:05
  • $\begingroup$ Did you downvote my answer Mr. @Eleven-Eleven? $\endgroup$ – Anastasiya-Romanova 秀 May 18 '14 at 13:18
  • $\begingroup$ 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 $1/n$? $\endgroup$ – André 3000 May 18 '14 at 14:13
  • $\begingroup$ @SpamIAm $$ \lim_{n\to\infty}\left[\left(1-\frac{1}{n^2}\right)^{ n^2}\right]^{\frac{1}{n}}=\left[\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n^2}\right]^{\lim_{n\to\infty}\frac{1}{n}} $$ $\endgroup$ – Anastasiya-Romanova 秀 May 18 '14 at 15:02
  • 1
    $\begingroup$ 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 $n \gg 1$. $\endgroup$ – Felix Marin Sep 4 '14 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.