# Limit of $\left(1-\frac{1}{n^2}\right)^n$

I'm trying to find the limit of $a_n = \left(1-\frac{1}{n^2}\right)^n$ for $n \rightarrow \infty$.

It seems that the limit is $1$, since $a_n = 0.999...$ for large $n$. The presentation $a_n = \frac{(n^2-1)^n}{n^{2n}}$ and expanding was my first idea, but I couldn't get the result from there. Any ideas?

• What do you need this limit for? One place where it crops up is if you define $e=\lim_{n\to\infty}(1+1/n)^n$ and then want to prove $\lim_{n\to\infty}(1-1/n)^n=1/e$, in which case some of the answers below will be not so useful. Nov 22, 2013 at 0:19
• Yes, all answers so far use properties of $log$ or $e$. How to figure it out without those properties? Nov 22, 2013 at 0:43
• Apr 21, 2014 at 6:02

$$\left(1-\frac1{n^2}\right)^n=\left(1-\frac1n\right)^n\left(1+\frac1n\right)^n\xrightarrow[n\to\infty]{}e^{-1}e=1$$

• @AdiDani $(ab)^n=a^nb^n$
– Pedro
Nov 22, 2013 at 0:35

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

• On a question asked two days ago, you beat me by 30 seconds :D +1 Nov 24, 2013 at 16:48
• I'd say this is "da proof"...+1 Nov 24, 2013 at 16:54
• Note also that you can prove the same way that $$\lim\left(1-\frac{1}{n^{1+\epsilon}}\right)^n=1$$ And I think that by looking at the reciprocal you also get $$\lim\left(1+\frac{1}{n^{1+\epsilon}}\right)^n=1$$ Nov 24, 2013 at 17:02

Since you asked for an answer not using the limit leading to $1/e$ how about this?

Using the binomial expansion of $\left( 1-\dfrac1{n^2} \right)^n$ you should be able to set up a series in $n$ and then take the limit. This series will start out $$1^n + n \cdot 1^{n-1} \cdot \dfrac{-1}{n^2} + \dfrac{n \cdot (n-1)}2 \cdot1^{n-2} \cdot \left( \dfrac{-1}{n^2}\right) ^2 \ldots$$ and from there you should be able to simplify and show that the limit goes to $1$.

• You need some minus signs around there.
– Pedro
Nov 22, 2013 at 1:26
• Thanks for catching that. I was so focused on the MathJax formatting that I forgot to include the sign. Nov 22, 2013 at 1:42
• Your idea of using the binomial theorem may not be as successful as expected: after all, the number of summands dependends on $\;n\;$ and thus we can not apply arithmetic of limits there. Nov 22, 2013 at 13:39

$$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty}\left(\left(1-\frac{1}{n^2}\right)^{n^{2}}\right)^{\frac{1}{n}}=\left(\frac{1}{e}\right)^{0}=1$$ because $$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n^{2}}=\frac{1}{e}$$ and $$\lim_{n\to\infty}{\frac{1}{n}}=0$$

• In general, passing to the limit only in part of the expression (while leaving the rest still with the variable tending to whatever) is wrong and can easily lead to serious mistakes. Nov 22, 2013 at 0:24
• This is not a correct solution.
– Pedro
Nov 22, 2013 at 0:29
• Is there a way proving this without using properties of $e$? Nov 22, 2013 at 0:44
• @AdiDani What is this law thou speaketh of?
– Pedro
Nov 22, 2013 at 2:14
• @DonAntonio Thanks a lot, I learned sth :) Nov 24, 2013 at 16:57