Question

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.

Answer 1

`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.