Question

How do you apply the ratio test to determine if `Sigma (5^n)/(6^n-5^n)` from `n=[1,oo)` is convergent to divergent?

Answer 1

The series:

`sum_(n=1)^oo 5^n/(6^n-5^n)`

is convergent.

Explanation:

Evaluate the ratio:

`abs(a_(n+1)/a_n) = abs ( (5^(n+1)/(6^(n+1)-5^(n+1)))/(5^n/(6^n-5^n)))`

`abs(a_(n+1)/a_n) = (5^(n+1)/5^n ) ( (6^n-5^n) / (6^(n+1)-5^(n+1)))`

`abs(a_(n+1)/a_n) = 5*6^n( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))`

`abs(a_(n+1)/a_n) = 5/6 *6^(n+1)( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))`

`abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / ((6^(n+1)-5^(n+1) )/ 6^(n+1)))`

`abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / (1-(5/6)^(n+1) ))`

Now as `5/6 < 1`

`lim_(n->oo) (5/6)^n = lim_(n->oo) (5/6)^(n+1) = 0`

so:

`lim_(n->oo) abs(a_(n+1)/a_n) = 5/6 lim_(n->oo) ( (1-(5/6)^n) / (1-(5/6)^(n+1) )) = 5/6 < 1`

which proves the series to be convergent.