1. ## converging with e

problem is: An = ( (e^2n)/(n^2 +3n - 1) )

my question is: Can i decide what number this problem is converging to by dividing the top and bottom of this equation by the largest n in the denominator? By doing that, what does this determine? just a guess?

Also, I'm unsure how to convert the natural log to something I can work with.

could i do this first --> 2n ln(e) --> 2n (1) --> so 2n?

this problem diverges, btw.

-thanks for any help.

2. Originally Posted by rcmango
problem is: An = ( (e^2n)/(n^2 +3n - 1) )

my question is: Can i decide what number this problem is converging to by dividing the top and bottom of this equation by the largest n in the denominator? By doing that, what does this determine? just a guess?

Also, I'm unsure how to convert the natural log to something I can work with.

could i do this first --> 2n ln(e) --> 2n (1) --> so 2n?

this problem diverges, btw....

Hello,

this is only a try, maybe it helps nevertheless:

$\lim_{n \rightarrow \infty}{\left( \frac{e^{2n}}{n^2+3n-1} \right)}$ simplifies for very large number n to:

$\lim_{n \rightarrow \infty}{\left( \frac{e^{2n}}{n^2} \right)}=\lim_{n \rightarrow \infty}{\left( \frac{e^{n}}{n} \right)^2} = \left( \lim_{n \rightarrow \infty}{ \frac{e^{n}}{n}} \right)^2$

And as you may know the limit $\lim_{n \rightarrow \infty}{ \frac{e^{n}}{n}}$ doesn't exist, that means $A_n$ diverges.

EB

3. Originally Posted by rcmango
problem is: An = ( (e^2n)/(n^2 +3n - 1) )

my question is: Can i decide what number this problem is converging to by dividing the top and bottom of this equation by the largest n in the denominator? By doing that, what does this determine? just a guess?

Also, I'm unsure how to convert the natural log to something I can work with.

could i do this first --> 2n ln(e) --> 2n (1) --> so 2n?

this problem diverges, btw.

-thanks for any help.
The thing you need to know to make this easy is that $e^x$ increases faster than any polynomial in $x$, so $A_n \to \infty$ as $n \to \infty$.

Alternativly you can use L'Hopitals's rule twice to discover the same result.

RonL