Limit from n=1 to n=infinity?

Question

3^n + ^6^n
-----------------
10^n

hmm... i should rephrase ...- im looking for the sum of the function from n=1 to n=infinity

Answer 1

You have (3/10)^n + (6/10)^n. Both are geometric series and since both fractions are less than 1, both sum to a finite amount. In case you forgot the sum of a geometric series, you can derive it by assuming S is finite in the equation 1 + a^1 + ... = S and note that a*(S - 1) = S, and thus S = a/(1-a). In your case, just subtract 1 from the sum, since you start at n=1.

Answer 2

The limit as n goes to infinity is zero.

For sufficiently large n, the 3^n is irrelevant compared to the 6^n. So, you have (6/10)^n, which clearly goes to zero as n goes to infinity.