Limit of $\left(1+\frac{1}{\sqrt n}\right)^n$ as n tends to $\infty$

Question

I'm trying to find $\lim_{n \to \infty}{\left(1+\frac{1}{\sqrt n}\right)^n}$. It looks pretty easy, so I think there's probably something simple I'm missing, because I can't seem to figure out where to even start. I've tried some basic algebraic manipulation but it doesn't seem to help much. I know the sequence diverges to infinity from typing it into WolframAlpha, but I don't know how to show that it's unbounded.

I'm trying to do it without using L'Hospital's rule or Bernoulli's inequality. Other Calc 1 methods are fair game.

Answer 1

Here a way without using Bernoulli directly but using the binomial formula only: $\left(1+\frac{1}{\sqrt n}\right)^n= 1 + \binom{n}{1}\frac{1}{\sqrt n} + \binom{n}{2}\left(\frac{1}{\sqrt n}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{\sqrt n}\right)^n \stackrel{n>1}{>} 1 + \binom{n}{1}\frac{1}{\sqrt n} = 1+\sqrt{n}$

Answer 2

Note that

$$\left(1+\frac{1}{\sqrt n}\right)^n=\left[\left(1+\frac{1}{\sqrt n}\right)^{\sqrt n}\right]^{\frac n {\sqrt n}}\ge2^{\frac n {\sqrt n}}\to \infty$$

or as an alternative

$$\left(1+\frac{1}{\sqrt n}\right)^n=e^{n\log\left(1+\frac{1}{\sqrt n}\right)}\sim e^{\frac n {\sqrt n}-\frac12}\to \infty$$

Answer 3

As $n\to\infty$, $$\left(1+\frac1{\sqrt n}\right)^{\sqrt n}\to e$$ and $e^{\sqrt n}$ is divergent since $e>1$.

Answer 4

Bernoulli's Inequality gives

$$\left(1+\frac1{\sqrt n}\right)^n\ge1+\sqrt n\to \infty$$