## anonymous 5 years ago series n!/n^n converge or diverge? how do you find out the convergence of ((2n^3+5n^2+3)/(4n^3+n+7))^n?

1. anonymous

for second one try to apply root test and then factor n^3 out and then take limit.

2. anonymous

ok i understood that. is there any other way i can do this

3. anonymous

existence, you should note that$\frac{n!}{n^n}-\frac{n}{n}.\frac{n-1}{n}.\frac{n-2}{n}.....\frac{3}{n}\frac{2}{n}\frac{1}{n}$As n goes to infinity, the left-hand numbers become asymptotically equivalent to 1, while the right-hand numbers approach zero, so you end up with a situation where$\frac{n!}{n^n} \iff 1.1.1.....0.0.0=0$where the if and only if sign is meant to mean asymptotic behavior.

4. anonymous

For the last one, you have that$2n^3$is asymptotically equivalent to $2n^3+5n^2+3$(i.e.for n large, they behave the same way) and for $4n^3+n+7$you have that this is asymptotically equivalent to $4n^3$As a consequence, for large n, you have$\left( \frac{2n^3+5n^2+3}{4n^3+n+7} \right)^n \approx \left( \frac{2n^3}{4n^3} \right)^n=\left( \frac{1}{2} \right)^n=\frac{1}{2^n}$so as n tends to infinity, your expression tends to 0 since 1/2^n tends to zero. I've included some information on asymptotic equivalence.

5. anonymous

thank you ver much

6. anonymous

Thank me by becoming a fan :) Took forever to write out :p And you're welcome.