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?

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 $n \rightarrow \infty$.
