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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?

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    $\begingroup$ 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. $\endgroup$ Nov 22, 2013 at 0:19
  • $\begingroup$ Yes, all answers so far use properties of $log$ or $e$. How to figure it out without those properties? $\endgroup$ Nov 22, 2013 at 0:43
  • $\begingroup$ See also: math.stackexchange.com/questions/762625/… $\endgroup$ Apr 21, 2014 at 6:02

4 Answers 4

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

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  • $\begingroup$ @AdiDani $(ab)^n=a^nb^n$ $\endgroup$
    – Pedro
    Nov 22, 2013 at 0:35
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By Bernoulli's inequality

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

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  • $\begingroup$ On a question asked two days ago, you beat me by 30 seconds :D +1 $\endgroup$
    – N. S.
    Nov 24, 2013 at 16:48
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    $\begingroup$ I'd say this is "da proof"...+1 $\endgroup$
    – DonAntonio
    Nov 24, 2013 at 16:54
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    $\begingroup$ 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 $$\lim\left(1+\frac{1}{n^{1+\epsilon}}\right)^n=1$$ $\endgroup$
    – N. S.
    Nov 24, 2013 at 17:02
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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 $1$.

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  • $\begingroup$ You need some minus signs around there. $\endgroup$
    – Pedro
    Nov 22, 2013 at 1:26
  • $\begingroup$ Thanks for catching that. I was so focused on the MathJax formatting that I forgot to include the sign. $\endgroup$ Nov 22, 2013 at 1:42
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Nov 22, 2013 at 13:39
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$$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^n=\lim_{n\to\infty}\left(\left(1-\frac{1}{n^2}\right)^{n^{2}}\right)^{\frac{1}{n}}=\left(\frac{1}{e}\right)^{0}=1$$ because $$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n^{2}}=\frac{1}{e}$$ and $$\lim_{n\to\infty}{\frac{1}{n}}=0$$

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    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Nov 22, 2013 at 0:24
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    $\begingroup$ This is not a correct solution. $\endgroup$
    – Pedro
    Nov 22, 2013 at 0:29
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    $\begingroup$ Is there a way proving this without using properties of $e$? $\endgroup$ Nov 22, 2013 at 0:44
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    $\begingroup$ @AdiDani What is this law thou speaketh of? $\endgroup$
    – Pedro
    Nov 22, 2013 at 2:14
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    $\begingroup$ @DonAntonio Thanks a lot, I learned sth :) $\endgroup$
    – hhsaffar
    Nov 24, 2013 at 16:57

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