You are here:

Advanced Math/convergence of series

Advertisement

Expert: - 11/22/2011


Question
How do you prove by comparison test that n!/(n^3) converges or diverges?

Answer
Compare n!/n^3   to 1/n

n!/n^3 > 1/n

(n-1)!/n^2 > 1/n

(n-1)!/n > 1

(n-2)!  (n-1)/n  > 1

Since (n-1)/n goes to 1 as n goes to plus infinity,
and (n-2)! goes to plus infinity , the left side goes to plus infinity and is thus greater than 1 for sufficiently large n.

It follows that n!/n^3 > 1/n for sufficiently large n and sum n!/n^3 diverges by the comparison test because sum 1/n diverges  

Advanced Math

All Answers

Answers by Expert:

Ask Experts

Volunteer

Socrates

Expertise

I can answer any questions from the standard four semester Calulus sequence. Derivatives, partial derivatives, chain rule, single and multiple integrals, change of variable, sequences and series, vector integration (Green`s Theorem, Stokes, and Gauss) and applications. Pre-Calculus, Linear Algebra and Finite Math questions are also welcome.

Experience

Ph.D. in Mathematics and many years teaching undergraduate courses at three state universities.

Education/Credentials
B.S. , M.S. , Ph.D.

©2012 About.com, a part of The New York Times Company. All rights reserved.