Converging sequence

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July 19th 2012, 09:54 AM
Ragnarok

Converging sequence

Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its limit.

$a_n = \left(1 + \frac{1}{n^2}\right)^n$

I was never very good at limits. I know that the limit of $\left(1 + \frac{1}{n}\right)^n$ 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!
July 19th 2012, 10:40 AM
girdav

Re: Converging sequence

Hint: take logarithm and use for $x\geq 0$ : $0\leq \ln(1+x)\leq x$ .
July 19th 2012, 10:59 AM
skeeter

Re: Converging sequence

let $y = \lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n$

$\ln{y} = \lim_{n \to \infty} \ln \left(1 + \frac{1}{n^2}\right)^n$

$\ln{y} = \lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n^2}\right)$

$\ln{y} = \lim_{n \to \infty} \frac{\ln \left(1+\frac{1}{n^2}\right)}{\frac{1}{n}}$

use L'Hopital for the indeterminate form $\frac{0}{0}$ ...

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

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

$y = e^0 = 1$
July 21st 2012, 07:21 AM
Ragnarok

Re: Converging sequence

Thank you!

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