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.
 A: There is an easy inductive proof that $(1+{1\over N})^N\le3-{1\over N}$ for all $N\in\mathbb{N}$, from which $(1+{1\over N})^N\lt3$ for all $N$ follows.  Letting $N=n^2$, this tells us that
$$\left(1+{1\over n^2}\right)^n\lt3^{1/n}$$
If you now prove that $3^{1/n}\to1$ as $n\to\infty$, the Squeeze Theorem says that $(1+{1\over n^2})^n\to1$ as well.
A: One of the most useful inequalities in asymptotic analysis is $$e^x \ge 1+x$$ (with near equality for small $x$, and equality for $x=0$). Here, it tells us that $1 + \frac1{n^2} \le e^{1/n^2}$, so $\left(1 + \frac1{n^2}\right)^n \le e^{1/n}$. As $n \to \infty$, $\frac1n \to 0$ and $e^{1/n} \to 1$; on the other hand, $\left(1+\frac1{n^2}\right)^n > 1$ for all $n$, so we're done by the squeeze theorem.
(In general, the inequality $e^x \ge 1+x$ helps us turn everything into a product while maintaining an upper bound that's fairly tight when $x$ is small.)
A: Here is a purely algebraic approach.  First we note that
$$\left(1+\frac1{n^2}\right)\left(1-\frac1{n^2}\right)\le 1$$
From $(1)$ it is easy to see that
$$\left(1+\frac1{n^2}\right)^n\le \frac1{\left(1-\frac{1}{n^2} \right)^n}\tag 2$$
Applying Bernoulli's Inequality to the term on the left-hand side of $(2)$ reveals 
$$1\le \left(1+\frac1{n^2}\right)^n\le \frac1{1-\frac1n}$$
whereupon applying the sqeeze theorem yields the covetes limit
$$ \lim_{n\to\infty} \left(1+\frac1{n^2}\right)^n=1$$
A: You could have just committed to your original idea:
$$\lim_{n\to\infty} \left(1+\frac{1}{n^2}\right)^n = \lim_{n\to\infty}\left(1+\frac{i}{n}\right)^n\cdot \lim_{n\to\infty}\left(1-\frac{i}{n}\right)^n
= e^i \cdot e^{-i} = 1$$
A: 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 $1$.
A: Use equivalents from the following equality:
$$a_n=\exp\left(n\ln\left(1+\frac{1}{n^2}\right)\right).$$
Recall that $\ln(1+x)=x+o(x)$ and that $\exp$ is continuous.
A: Simple:
$$a_n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\stackrel{n\to\infty}\longrightarrow e^{1/\infty}=e^0=1$$
Using the historically first definition of $e$:
$$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}$$
A: If $n \to \infty$ the this becomes form of $1^{\infty}$ to which you get, $e^{\lim_{n \to \infty} \frac{1}{n}}$ and you get $e^{0}$ which is 1.
