Prove that $( 1 + n^{-2}) ^n \to 1$.

I need to prove that $\left(1 + \frac 1 {n^2} \right)^n \to 1$.

I tried to use Bernoulli's inequality, but that is not very useful since in the original sequence there is a plus sign. I then tried to use the Sandwich Theorem by finding two sequences which would make bounds for the original one. The lower bound is obvious, the upper bound not so much. I tried using the sequence $\left( \frac 1 {n+1} \right) ^{\frac 1 n}$, but I could not show that this sequence is bigger than the original one for all $n$. Could anyone help me with this?

Solution 1:

One may write, as $n \to \infty$, $$ \left(1+\frac1{n^2} \right)^n=e^{\large n\log\left(1+\frac1{n^2} \right) }\sim e^{\large n\times\frac1{n^2} }=e^{\large \frac1{n}} \to 1. $$

Solution 2:

Use $e^x \geq 1 + x$ for all $x \in \mathbb{R}$. There are many ways to see this. The easiest, in my opinion, is to note that $y = x + 1$ is tangent to $y = e^x$ and $x \mapsto e^x$ is convex, then use definition of convexity.

It follows that $$ 1 \leq \left(1 + \frac{1}{n^2}\right)^n \leq \left(\exp\frac{1}{n^2}\right)^n = \exp\frac{1}{n} \rightarrow 1 $$ as $n \rightarrow \infty$.

Solution 3:

Using Taylor approximation to $\log(1 + x)$ for $x \to 0$ gives $$ (1+n^{-2})^{n} = \exp [ n \log (1 + n^{-2})] = \exp [n\cdot n^{-2} + n\cdot n^{-2}\cdot o(1)] = \exp \bigg[ \frac{1}{n} + n^{-1}o(1) \bigg] \to e^{0} = 1 $$ as $n \to \infty$.