Question

?How do you find the sum of the infinite geometric series `6^n/7^n`?

Answer 1

`7`

Explanation:

`(x^(n+1)-1)/(x-1)=1+x+x^2+cdots+x^n` then

`sum_(k=0)^n (6/7)^k = ((6/7)^(n+1)-1)/((6/7)-1)`

Now,

`lim_(n->oo)sum_(k=0)^n (6/7)^k=lim_(n->oo) ((6/7)^(n+1)-1)/((6/7)-1)=1/(1-6/7)` because

`lim_(n->oo)(6/7)^n=0` so finally

`sum_(k=0)^oo (6/7)^k = 7`