Convergence of a very simple series

Click For Summary
SUMMARY

The series \(\sum\limits_{n = 2}^\infty {\frac{{n^2 }}{{1 - n^3 }}}\) diverges. The term \(a_n = \frac{n^2}{1 - n^3}\) is always negative due to a positive numerator and a negative denominator. The user initially considered using the comparison test and absolute convergence test but found them unhelpful. The discussion concludes that rewriting the series as \(S = -s\) where \(s = \sum_{n=2}^{\infty}\frac{n^{2}}{n^{3}-1}\) allows for further analysis of convergence or divergence.

PREREQUISITES
  • Understanding of series convergence tests, including the comparison test and absolute convergence test.
  • Familiarity with series notation and manipulation, particularly with infinite series.
  • Knowledge of polynomial behavior in limits, specifically how to analyze terms like \(n^2\) and \(n^3\).
  • Basic algebraic manipulation skills to rewrite series for analysis.
NEXT STEPS
  • Study the Ratio Test for series convergence to apply it to similar series.
  • Learn about the Root Test and its application in determining series convergence.
  • Explore the concept of conditional convergence and its implications in series analysis.
  • Investigate the behavior of series with negative terms and their convergence properties.
USEFUL FOR

Mathematicians, students studying calculus or real analysis, and anyone interested in series convergence and divergence analysis.

Benny
Messages
577
Reaction score
0
Hi, there's a really simple looking series that I don't know how to deal with.
[tex] \sum\limits_{n = 2}^\infty {\frac{{n^2 }}{{1 - n^3 }}} [/tex]
How would I determine whether or not this series converges by using some standard convergence tests? If a_n = (n^2)/(1-n^3) then the numerator is always positive while the denominator is always negative so that a_n is always negative. So I can't think of a way to use the comparison test, limit comparison test etc. Since a_n looks like 1/n I have a feeling that the series diverges. But I can't think of any tests to use to verify whether or not my hypothesis is correct. At first I thought about using the absolute convergence test but if a series is not absolutely convergent, it can still be convergent so that didn't really help.
Can someone help me with this one?

Note: I would like to be able to do this without the integral test if possible.
 
Last edited:
Physics news on Phys.org
Rewrite your series S as: [itex]S=-s, s=\sum_{n=2}^{\infty}\frac{n^{2}}{n^{3}-1}[/tex]<br /> S diverges or converges with s.[/itex]
 
Oh ok, thanks for the help arildno.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K