choosing n given epsilon

**BrainMan** · September 29th 2007, 07:42 AM

I asked this before and the guy said it was divergent; however, it's clearly not. I just wanted some help on how to do it.

I need to find an N(epsilon) for the following sequence such that for all n>=N(epsilon) the absolute value of the nth element is less than epsilon >0:

n(sqrt((n^4)+4) - n^2) (i.e. the n is multiplied to both the sqrt((n^4)+4)) and n^2 with subtraction between those two terms. I know that it converges to zero. I just don't know how to find a sequence (it must be strictly decreasing) that bounds it and makes finding the N(epsilon) easy.

I appreciate any help.

**red_dog** · September 29th 2007, 08:30 AM

$a_n=n(\sqrt{n^4+4}-n^2)=\frac{n(n^4+4-n^4)}{\sqrt{n^4+4}+n^2}=\frac{4n}{\sqrt{n^4+4}+n^2 }<\frac{4n}{n^2}=\frac{4}{n}<\epsilon$
Then $n>\frac{4}{\epsilon}$ .
So, take $n_{\epsilon}=\left[\frac{4}{\epsilon}\right]+1$

Thread: choosing n given epsilon

LinkBack

Thread Tools

Display

choosing n given epsilon

Similar Math Help Forum Discussions

Choosing

Solving Delta Epsilon Proof (Given Epsilon)

choosing a committee

Choosing the best ad

choosing the parameters

Search Tags