Convergence comparison test (sequence, calc II)

coals
Messages
3
Reaction score
0

Homework Statement


Finally done with stupid improper integrals (calc 2 over summer is hard work, moving super fast) and now I'm sequences and series and what have you. I have another comparison test problem

Instructions are to find out if sequence converges or diverges, and find limit if it converges.

Sequence is:
{An}= n!/(2^n) .

The Attempt at a Solution


I know what to do with the factorial most of the time. I tried expanding it but the 2^n is throwing me off becuase the sequence diverges. Can i get a nudge in the right direction? I don't think my instructor wants me to find the limit of it if i can avoid doing so on divergence problems (he makes a point to tell us on HW to check convergence beforehand as practice for quizzes/tests).

As always thank all of you who help out pity-able college students as myself (lol) .
 
Physics news on Phys.org
You should really show us what you tried. But try a ratio test.
 
Its a sequence or a series ?
There is no comparison test for sequences.
If its a sequence: Try the sandwich theorem.
If its a series: See Dick's reply.
 
a_(n+1) / a_(n) >= 1, so the sequence monotonic increasing.
if you can't find a limit (number) then the series goes to infinity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
3
Views
1K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
4
Views
1K
Replies
6
Views
1K
Replies
14
Views
2K
Back
Top