Question

How do you determine if `Sigma (7^n-6^n)/5^n` from `n=[0,oo)` converge and find the sums when they exist?

Answer 1

See below.

Explanation:

`(7^n-6^n)/5^n = (7/5)^n-(6/5)^n`

`( (7/5)^n-(6/5)^n)/(7/5-6/5)=(7/5)^(n-1)+(7/5)^(n-2)(6/5)+ cdots + (7/5)(6/5)^(n-2)+(6/5)^(n-1)` or

`(7/5)^n-(6/5)^n=1/5((7/5)^(n-1)+(7/5)^(n-2)(6/5)+ cdots + (7/5))`

Here `7/5 > 1` and `6/5 > 1` so for `n > 5` we surely will have

`1/5((7/5)^(n-1)+(7/5)^(n-2)(6/5)+ cdots + (7/5)) > 1`

Concluding this is a divergent series.