• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Convergent/Divergent Series

  • Thread starter zandbera
  • Start date
1. Homework Statement
I have to determine whether the given series is convergent or divergent using the comparison tests:
[tex]\sum[/tex] from n = 1 to infinity of (n + 4n / (n + 6n)


2. Homework Equations
If bn is convergent and an [tex]\leq[/tex] bn then an is also convergent

liimit of an/bn as n goes to infinity = c, if c > 0, then both are either convergent or divergent

3. The Attempt at a Solution

I tried saying that bn was (4/6)^n but i dont know how to compare that to the original series
 

HallsofIvy

Science Advisor
Homework Helper
41,682
864
Certainly [itex](n+4^n)/(n+ 6^n)< (n+ 4^n)/6^n[/itex] because the left side has a larger denominator. It is also true that n< 4^n for any positive integer n. That means that [itex]n+ 4^n< 4^n+ 4^n< 2(4^n)[/itex] and so [itex](n+4^n)/(n+6^n)< (n+4^n)/6^n< 2(4^n)/6^n)[/itex].
 
so then because 2(4/6)^n is convergent, the original series is convergent, correct?
 
1,356
0
Correct.
 

Related Threads for: Convergent/Divergent Series

  • Posted
Replies
2
Views
931
Replies
4
Views
6K
  • Posted
Replies
16
Views
2K
  • Posted
Replies
4
Views
732
  • Posted
Replies
4
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top