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I want to compute the limit $\left(1+\frac{1}{n^2}\right)^n$. One way to do this is to take logs. So $$x_n=\left(1+\frac{1}{n^2}\right)^n$$ Then $$\log x_n = n\log\left(1+\frac{1}{n^2}\right) = n\left(\frac{1}{n^2}-\frac{1}{2n^4}+O\left(\frac{1}{n^6}\right)\right)$$So $\log x_n$ converges to $0$, and $x_n$ converges to $1$.

Is there a simpler way to do this, maybe without Taylor log expansion?

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  • $\begingroup$ You could also use l'Hosptial's rule with $\log(1 + 1/n^2)/(1/n)$ but that's not really simpler. $\endgroup$ – Max Sherman Jun 9 '13 at 2:17
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From the Taylor Series, $$1+\frac{1}{n^2}\leq \exp\left(\frac{1}{n^2}\right)$$

Thus, $$1\leq \left(1+\frac{1}{n^2}\right)^n \leq \exp\left(\frac{1}{n}\right)$$ so by the Squeeze Theorem, the result follows $\square$

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  • $\begingroup$ Do you mean $\exp\left(\frac{1}{n^2}\right)$ in the last line? $\endgroup$ – PJ Miller Jun 9 '13 at 2:59
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    $\begingroup$ @PJMiller He raised both sides to the $n$, so $\exp(\frac{1}{n^{2}})^{n}=\exp(\frac{1}{n})$. $\endgroup$ – Jackson Walters Jun 9 '13 at 3:12
  • $\begingroup$ @JacksonWalters That's right, my mistake. $\endgroup$ – PJ Miller Jun 9 '13 at 3:15
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Claim: Let $\,\{a_n\}\,$ be a non-negative convergent sequence s.t.

$$\lim_{n\to\infty}a_n=a>0\;,\;\;\text{then}\;\;\lim_{n\to\infty}\sqrt[n]{a_n}=1$$

Proof: there exists $\,N\in\Bbb N\,$ s.t.

$$n>N\implies |a_n-a|<\frac a2\iff 0<\frac a2<a_n<\frac{3a}2\implies$$

$$\sqrt[n]\frac a2\le\sqrt[n]{a_n}\le\sqrt[n]\frac{3a}2$$

and since $\,\displaystyle{\sqrt[n]x\xrightarrow[n\to\infty]{}1\;\;\;\forall\,x>0}\,$ , we're done by the squeeze theorem.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$

Thus, applying the above lemma we get that

$$\left(1+\frac1{n^2}\right)^{n^2}\xrightarrow[n\to\infty]{}e>0\implies\left(1+\frac1{n^2}\right)^n=\left[\left(1+\frac1{n^2}\right)^{n^2}\right]^{\frac1n}\xrightarrow[n\to\infty]{}1$$

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    $\begingroup$ Also nice proof! $\endgroup$ – PJ Miller Jun 9 '13 at 13:40

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