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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?

Adi Dani
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Leif Sabellek
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    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. – hmakholm left over Monica Nov 22 '13 at 00:19
  • Yes, all answers so far use properties of $log$ or $e$. How to figure it out without those properties? – Leif Sabellek Nov 22 '13 at 00:43
  • See also: http://math.stackexchange.com/questions/762625/lim-n-to-infty-left1-frac-1n2-rightn – Martin Sleziak Apr 21 '14 at 06:02

4 Answers4

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

DonAntonio
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By Bernoulli's inequality

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

WimC
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  • On a question asked two days ago, you beat me by 30 seconds :D +1 – N. S. Nov 24 '13 at 16:48
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    I'd say this is "da proof"...+1 – DonAntonio Nov 24 '13 at 16:54
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    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 $$\lim\left(1+\frac{1}{n^{1+\epsilon}}\right)^n=1$$ – N. S. Nov 24 '13 at 17:02
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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 $$\lim_{n\to\infty}{\frac{1}{n}}=0$$

Adi Dani
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    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. – DonAntonio Nov 22 '13 at 00:24
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    This is not a correct solution. – Pedro Nov 22 '13 at 00:29
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    Is there a way proving this without using properties of $e$? – Leif Sabellek Nov 22 '13 at 00:44
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    @AdiDani What is this law thou speaketh of? – Pedro Nov 22 '13 at 02:14
  • It's a simple lemma to prove that if $;a_n\xrightarrow[n\to\infty]{}a;,;;\text{and};;0<a;$ , then $;\sqrt[n]{a_n}\xrightarrow[n\to\infty]{}1;$. This is based in $;\sqrt[n]a\xrightarrow[n\to\infty]{}1;$ for $;a>0;$. Yet, to evaluate part of the expression of a limit while leaving other part is wrong. – DonAntonio Nov 22 '13 at 05:57
  • @DonAntonio Very interesting point, can you give an example that doing this causes mistake? – hhsaffar Nov 24 '13 at 16:50
  • Of course, @hhsaffar. Let's take a very well known limit: $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty} 1^n=1;\ldots\ldots$$ – DonAntonio Nov 24 '13 at 16:54
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    @DonAntonio Thanks a lot, I learned sth :) – hhsaffar Nov 24 '13 at 16:57
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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 $1$.

half-integer fan
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  • You need some minus signs around there. – Pedro Nov 22 '13 at 01:26
  • Thanks for catching that. I was so focused on the MathJax formatting that I forgot to include the sign. – half-integer fan Nov 22 '13 at 01:42
  • 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. – DonAntonio Nov 22 '13 at 13:39