Is this Series Convergent or Divergent Using Comparison Tests?

[zandbera]

Homework Statement

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)

Homework Equations

If b_n is convergent and a_n \leq b_n then a_n is also convergent

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

The Attempt at a Solution

I tried saying that bn was (4/6)^n but i don't know how to compare that to the original series

 

[HallsofIvy]

Certainly (n+4^n)/(n+ 6^n)&lt; (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&lt; 4^n+ 4^n&lt; 2(4^n) and so (n+4^n)/(n+6^n)&lt; (n+4^n)/6^n&lt; 2(4^n)/6^n).

 

[zandbera]

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

 

[e(ho0n3]

Correct.