# Limit calculation Real Analysis (Calculus Results prohibited): $\lim_{n\to\infty} \left(1+\frac1{\sqrt{n}}\right)^n$ [duplicate]

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.

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