# Limits of Sequences

• Jul 12th 2012, 05:48 PM
Calctaker
Limits of Sequences
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.

an = [en + (-3)n] / 5n [solved]

an = en / 2n [solved]

an = [1+(1/n)]n

an = n[(n2 + 1)1/2 - n]

Any help would be greatly appreciated (Nod)
• Jul 12th 2012, 05:57 PM
richard1234
Re: Limits of Sequences
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.
• Jul 12th 2012, 07:08 PM
Reckoner
Re: Limits of Sequences
Quote:

Originally Posted by Calctaker
an = [1+(1/n)]n

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

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

Quote:

Originally Posted by Calctaker
an = n[(n2 + 1)1/2 - n]

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

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

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

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

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

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

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

$=\frac12$
• Jul 12th 2012, 07:29 PM
Calctaker
Re: Limits of Sequences
Thank you very very much!!!!!!