1. ## 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!

2. ## Re: Converging sequence

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

3. ## 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 $\displaystyle \frac{0}{0}$ ...

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{-\frac{2}{n^3}}{1 + \frac{1}{n^2}}} \cdot -\frac{n^2}{1}}$

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\frac{2}{n}}{1 + \frac{1}{n^2}}} = 0$

$\displaystyle y = e^0 = 1$

Thank you!