
Converging sequence
Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its limit.
$\displaystyle a_n = \left(1 + \frac{1}{n^2}\right)^n$
I was never very good at limits. I know that the limit of $\displaystyle \left(1 + \frac{1}{n}\right)^n$ as n goes to infinity is e, but only because they made us memorize it. I don't understand why that is, or how to use it to solve this problem. About the only technique I know you can use is the squeeze theorem, but I never understood how to come up with the two squeezing functions. Could someone explain how to do this problem?
Thank you!

Re: Converging sequence
Hint: take logarithm and use for $\displaystyle x\geq 0$: $\displaystyle 0\leq \ln(1+x)\leq x$.

Re: Converging sequence
let $\displaystyle y = \lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n$
$\displaystyle \ln{y} = \lim_{n \to \infty} \ln \left(1 + \frac{1}{n^2}\right)^n$
$\displaystyle \ln{y} = \lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n^2}\right)$
$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\ln \left(1+\frac{1}{n^2}\right)}{\frac{1}{n}}$
use L'Hopital for the indeterminate form $\displaystyle \frac{0}{0}$ ...
$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\frac{2}{n^3}}{1 + \frac{1}{n^2}}} \cdot \frac{n^2}{1}}$
$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\frac{2}{n}}{1 + \frac{1}{n^2}}} = 0$
$\displaystyle y = e^0 = 1$

Re: Converging sequence