# Convergent/Divergent Series

1. Mar 30, 2009

### zandbera

1. The problem statement, all variables and given/known data
I have to determine whether the given series is convergent or divergent using the comparison tests:
$$\sum$$ from n = 1 to infinity of (n + 4n / (n + 6n)

2. Relevant equations
If bn is convergent and an $$\leq$$ 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

2. Mar 30, 2009

### HallsofIvy

Staff Emeritus
Certainly $(n+4^n)/(n+ 6^n)< (n+ 4^n)/6^n$ because the left side has a larger denominator. It is also true that n< 4^n for any positive integer n. That means that $n+ 4^n< 4^n+ 4^n< 2(4^n)$ and so $(n+4^n)/(n+6^n)< (n+4^n)/6^n< 2(4^n)/6^n)$.

3. Mar 30, 2009

### zandbera

so then because 2(4/6)^n is convergent, the original series is convergent, correct?

4. Mar 30, 2009

Correct.