Limits of Sequences

**Calctaker** · July 12th 2012, 05:48 PM

Hey guys,
I am struggling with the following homework problems:

use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges.

a_n = [eⁿ + (-3)ⁿ] / 5ⁿ [solved]

a_n = eⁿ / 2ⁿ [solved]

a_n = [1+(1/n)]ⁿ

a_n = n[(n² + 1)^1/2 - n]

Any help would be greatly appreciated

**richard1234** · July 12th 2012, 05:57 PM

The first one converges. Realize that $a_n = \frac{e^n}{5^n} + \frac{(-3)^n}{5^n} = (\frac{e}{5})^n + (-\frac{3}{5})^n$ , both sequences converge to zero.

**Reckoner** · July 12th 2012, 07:08 PM

Originally Posted by Calctaker

a_n = [1+(1/n)]ⁿ

$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$

This is one of the basic representations of $e.$

Originally Posted by Calctaker

a_n = n[(n² + 1)^1/2 - n]

$\lim_{n\to\infty}n\left(\sqrt{n^2+1}-n\right)$

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

$=\lim_{n\to\infty}\frac{n\left[\left(n^2+1\right)-n^2\right]}{\sqrt{n^2+1}+n}$

$=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}+n}$

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

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

$=\lim_{n\to\infty}\left(\sqrt{1+\frac1{n^2}}+1 \right)^{-1}$

$=\frac12$

**Calctaker** · July 12th 2012, 07:29 PM

Thank you very very much!!!!!!

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