# Convergent/Divergent Series

#### zandbera

1. 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)

2. Homework 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

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#### HallsofIvy

Homework Helper
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 $(n+4^n)/(n+6^n)< (n+4^n)/6^n< 2(4^n)/6^n)$.

#### zandbera

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

Correct.

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