How do you use the direct comparison test to determine if Sigma 3^n/(4^n+5) from [0,oo) is convergent or divergent?

1 Answer
Feb 1, 2018

See explanation.

Explanation:

4^n + 5 > 4^n which means that 1/(4^n+5) < 1/4^n.

Since 3^n >0, 3^n/(4^n+5) < 3^n/4^n for all n>=0.

Given this, we know sum3^n/(4^n+5) < sum3^n/4^n.

sum(3/4)^n is a convergent geometric series.

Since sum3^n/(4^n+5) is positive termed and sum3^n/(4^n+5) < sum3^n/4^n, sum3^n/(4^n+5) must converge by direct comparison.