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?

We can do this using just the binomial expansion, without involving $e$. Note that if $r \ge 1$, then
$$\binom{n}{r} = \dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r(r-1)(r-1)\cdots 2\cdot 1} \le n(n-1)(n-2)\cdots(n-r+1)\le n^r$$
and it's true for $r=0$, too, because $1 \le 1$. So the binomial theorem gives
$$\left(1+\frac{1}{n^2}\right)^n = \sum_{r=0}^n\binom{n}{r}\frac{1}{n^{2r}} \le \sum_{r=0}^n \frac{1}{n^r} < \sum_{r=0}^\infty \frac{1}{n^r} = \frac{1}{1-1/n}
$$
which tends to $1$ as $n \to\infty$.
