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?

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 $n \to \infty$.
