I want to determine the limit of the sequence: $$a_n=(1+ \frac{1}{n^2})^n \rightarrow 1$$ I was thinking that I could do this via the squeeze theorem, I definitely know that: $$ 1=(1)^n \leq (1+ \frac{1}{n^2})^n \leq (1 + \frac{1}{n})^n$$ Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?
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1Let $x_n=1+(1/n^2)$ then $n(x_n-1)\to 0 $ and hence $x_n^n\to 1$. See https://math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}\to 1,(1-n^{-2})^n\to 1$ and divide. – Paramanand Singh Nov 25 '18 at 16:11
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We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = \left(1+\frac{1}{n}\right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = \sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n \to 1$ and is consistent with what you did.
DeepSea
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Note that: $$\left(1+\frac1{n^2}\right)^n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\sim e^{1/n}\to_{\infty} 1.$$
farruhota
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You could also notice that by expansion,
$$\lim_{n \to \infty}\bigg(1+\frac{1}{n^2}\bigg)^n = \lim_{n \to \infty} \bigg[ {n \choose 0}+{n \choose 1}\bigg(\frac{1}{n^2}\bigg)+{n \choose 2}\bigg(\frac{1}{n^2}\bigg)^2+…$$
$$= \lim_{n \to \infty} \bigg[1+\frac{n!}{1!(n-1)!n^2}+\frac{n!}{2!(n-2)!n^4}+…$$
$$= \lim_{n \to \infty}\bigg[1+\frac{n}{n^2}+\frac{n(n-1)}{2 n^4}+…$$
And all the terms with $n$ vanish as $n \to \infty$ since the denominator dominates, leaving $1$.
KM101
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