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.
What I need help with is how do I prove that
I used the Limit Comparison Test
let
I was supposed to show a reason how
1 Answer
Feb 6, 2018
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
Hence