Given that sum from n=1 to infinity ( 3^n + 2^n) / 6^n.
Was asked to determine if it's convergent or divergent. If convergent, find its sum.
What i did was
(3/6)^n + (2/6)^n
= (1/2)^n + (1/3)^n
since both 1/2 and 1/3 are less than 1. so it's convergent.
hence I used a/1-r to find the sum to infinity for both 1/2 and 1/3 seperate cases and added up both the sum. But it was wrong. why?
Critical point: notice that the sums, as shown by Prove It: and , start at n= 1. The formula has the sum starting at n= 0. Prove It handled that by taking a= 1/3 and 1/2 so they were and .
You could also do that by noting that is just missing the "n= 0" term which is 1. So just use the standard formula, with a= 1, r= 1/3 or 1/2, and subtract 1:
serenalee, if you got instead of , that was your error.
(In the future, if you are asking why your answer was wrong, it would be a good idea to tell us what your answer was!)