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