Comparison Test?

fastcarslaugh · #1 $Old$ 11-19-2008, 06:45 PM

For each sequence $a_n$ find a number r such that
$\frac{a_n}{r^n}$ has a finite non-zero limit.

(This is of use, because by the limit comparison test the series $\displaystyle \sum_{n=1}^\infty a_n$ and $\displaystyle \sum_{n=1}^\infty r^n$ both converge or both diverge.)

$a_n = (4 + 7 ^ n)^{- 2}$

Can you please explain this question using that example?

Mathstud28 · #2 $Old$ 11-19-2008, 06:48 PM

Quote:

Originally Posted by fastcarslaugh $View Post$

For each sequence $a_n$ find a number r such that
$\frac{a_n}{r^n}$ has a finite non-zero limit.

(This is of use, because by the limit comparison test the series $\displaystyle \sum_{n=1}^\infty a_n$ and $\displaystyle \sum_{n=1}^\infty r^n$ both converge or both diverge.)

$a_n = (4 + 7 ^ n)^{- 2}$

Can you please explain this question using that example?

Do you see that as n gets really big that $\frac{1}{\left(4+7^n\right)^2}\approx\frac{1}{\left(7^n\right)^2}$ ? So we can say that as n goes to infinity $\frac{1}{(4+7^n)^2}\sim\frac{1}{14^n}$ . In other words $14^n$ dominates the denominator. Does that make sense?

Moo · #3 $Old$ 11-20-2008, 12:34 PM

Quote:

Originally Posted by Mathstud28 $View Post$

So we can say that as n goes to infinity $\frac{1}{(4+7^n)^2}\sim\frac{1}{14^n}$ .

Eh what ????
49 would be more logic huh?

Quote:

In other words $14^n$ dominates the denominator. Does that make sense?

not for me

What does it mean that it dominates the denominator ? I thought dominating was majoring ?
Though it's rather minoring here.

Geometric series.

fastcarslaugh · #4 $Old$ 11-20-2008, 05:44 PM

I still don't understand how you came up with the $14^n$

and what would r equal to?

Mathstud28 · #5 $Old$ 11-20-2008, 07:22 PM

Quote:

Originally Posted by fastcarslaugh $View Post$

I still don't understand how you came up with the $14^n$

and what would r equal to?

Consider this: if you have the functions $f(n)=7^n+n$ and $g(n)=7^n$ we can see for example that $\left|f(1)-g(1)\right|$ is significant (i.e. there is a real difference between them). But what about $\left|f(50000)-g(50000)\right|$ now we see that this is getting pretty small, so we can make the jump to say that for sufficient large n that $\left|f(n)-g(n)\right|\approx{0}$ . So they approach each other as n goes to infinity. The reason they approximate each other is think about it, if n gets really large how much bigger is $7^n$ then $n$ ? It gets much much larger! So the $n$ starts to become insignificant. So we can say that as n gets sufficiently large $7^n+n\sim{7^n}$ . In other words the $7^n$ "dominates" the $n$ , so we dont have to consider it for really large n. Now lets see if you can do it now. This might help, the notation $f(n)\succ{g(n)}$ means $f(n)$ dominates $g(n)$ . Well this will help you

$c\prec{n}\prec\cdots\prec{n^{10}}\prec\cdots\prec\left(a>1\right)^n$ . So as n goes to infinity you only have to consider the most dominant function.