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Finf $\lim_{n\to\infty} a_n$ where $$ a_n=\left(1+\frac1{\sqrt{n}}\right)^n$$

I was rationalizing the expression or the inverse of the expression and seeing where that leads to but I am stuck.

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Put $m = \sqrt{n} \implies L = \lim_{m \to \infty} \left( \left( 1+ \frac{1}{m} \right)^m \right)^m \ge \lim_{ m \to \infty } 2^m = \infty \implies L = \infty $ .

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  • $\begingroup$ Should $3^m$ be $2^m$? $\endgroup$ – saulspatz Mar 18 at 1:42
  • $\begingroup$ I know it’s either $2$ or $3$ but more than $1$ and constant. $\endgroup$ – DeepSea Mar 18 at 1:45
  • $\begingroup$ Is the proof of ((1+(1/m)^m)^m) > 2^m trivial? $\endgroup$ – Snoop Dogg Mar 18 at 1:47
  • $\begingroup$ @JohnDoe $\left(1+\frac1m\right)^m>2$ by the binomial theorem. $\endgroup$ – saulspatz Mar 18 at 1:48
  • $\begingroup$ @DeepSea It's not $3$, because $(1+1/m)^m\to e<3$ $\endgroup$ – saulspatz Mar 18 at 1:49
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$a_n=\left(1+\frac1{\sqrt{n}}\right)^n>n(\frac{1}{\sqrt n})\to \infty.$

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By binomial expansion we have $$\left(1+\frac{1}{\sqrt{n}}\right)^n=1+\frac{n}{\sqrt{n}}+\frac{n(n-1)}{2n}+\frac{n(n-1)(n-2)}{6 n \sqrt{n}}+...$$ $$=1+\sqrt{n}+\frac{n}{2}+O(n^{3/2})$$ So $$\lim_{n \to \infty} \left(1+\frac{1}{\sqrt{n}}\right)^n=\infty$$

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