A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Determine the convergence or divergence of the series. (2n)!/[(n1)3^n]
anonymous
 5 years ago
Determine the convergence or divergence of the series. (2n)!/[(n1)3^n]

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Diverge, because n! has a higher order than b^n for any constant b.

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean \[\frac{n!}{(n1)3^n}\] or \[\frac{n!}{(n1)^{3n}}\] ?

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Either of those is sequence, not a series. A sequence effectively a list of numbers. A series tells you to "add up all the numbers in the sequence."

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Regardless of whether they're sequences or series, the nth terms don't converge anyway. I didn't notice the second interpretation. Hmm...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.