
#1
Nov3012, 03:29 PM

P: 2

After doing my homework on testing for convergence/divergence in infinite series, I noticed that if you are testing for divergence of a rational function, if the difference in the degree of the functions (bottom  top) [itex]\leq[/itex] 1, then it is divergent, and if the difference in the degree of the functions is > 1, then it is convergent. I can post this proof if someone wants.
I was just wondering if there is a name for this concept.. 



#2
Nov3012, 03:38 PM

Sci Advisor
P: 5,935

From your question (as best as I can understand it), it seems to be a description of the ratio test.




#3
Nov3012, 03:42 PM

P: 2

It's similar to the ratio test, but all the ratio test says is that if the limit comes out to a positive finite number, then both of the series either diverge, or both converge. I used the ratio test in proving this, but this goes a step further and says that if the difference in degrees is less than or equal to 1, it diverges no matter what. And if it is greater than 1, it converges no matter what.




#4
Dec112, 08:59 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Question about divergence
No, what you give is the "comparison test", not the "ratio test".




#5
Dec112, 10:22 AM

P: 136




Register to reply 
Related Discussions  
a question that uses divergence thm  Calculus & Beyond Homework  2  
Why the divergence of a diagram when superficial degree of divergence D=0 is Ln(lambd  Quantum Physics  2  
Divergence Question  Calculus & Beyond Homework  1  
Question on Divergence  Calculus  5 