Easy way to find out limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$? What's an easy way to find out the limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$?
I don't think binomial expansion like with  $\left (1-\frac{1}{n^2} \right )^n = \left (1+\frac{1}{n} \right )^n \cdot \left (1-\frac{1}{n} \right )^n$ is possible.
And Bernoulli's inequality only shows $\left (1+\frac{1}{n^2} \right )^n \geq 1 + n \cdot \frac{1}{n^2} = 1 + \frac{1}{n} \geq 1$ which doesn't seem to help as well.

Using the binomial theorem and the fact that $\binom{n}{k}\leq\frac{n^k}{k!}$, we have: $$1\leq\left(1+\frac1{n^2}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac1{n^{2k}}\leq\sum_{k=0}^n\frac{n^k}{k!n^{2k}}=\sum_{k=0}^n\frac{1}{k!n^k}\leq1+\frac1n+\sum_{k=2}^n\frac1{n^2}\leq1+\frac2n.$$ Therefore, by the squeeze theorem, the limit is $1$.
