Test series for convergence/divergence. sum3^n/(4^n+5^n)?

What I need help with is how do I prove that b_n is convergent.

sum3^n/(4^n+5^n)
I used the Limit Comparison Test
a_n > b_n
3^n/(4^n+5^n) > 3^n/(5^n+5^n), n>=1
let n=1, 3/9 > 3/10

sumb_n is converges and a_n > b_n, so suma_n diverges

I was supposed to show a reason how sumb_n converges, but I'm not sure what to put down.

1 Answer
Feb 6, 2018

sum 3^n/(4^n+5^n) converges

Explanation:

Given:

a_n = 3^n/(4^n+5^n)

Note that:

0 < a_n = 3^n/(4^n+5^n) < 3^n/(2*4^n) = 1/2(3/4)^n

So:

0 < sum a_n < sum 1/2(3/4)^n

This latter sum converges, since it is a geometric series with common ratio 3/4 smaller than 1.

Hence sum a_n converges.