SOLUTION: sum from n = 1 to n = infinity(2^n - 1)/3^n

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Question 1118782: sum from n = 1 to n = infinity(2^n - 1)/3^n
Found 2 solutions by solver91311, greenestamps:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Depends. Do you mean



or




John

My calculator said it, I believe it, that settles it

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


S = (2^1-1)/3^1 + (2^2-1)/3^2 + {2^3-1/3^3 + ...

S = 1/3 + 3/9 + 7/27 + 15/81 + ...

If the denominators of the series form a geometric progression but the whole series is not geometric, then multiply that expression for the sum by the common ratio; then subtract the two series.
  3S = 1 + 3/3 + 7/9 + 15/27 + ...
   S =     1/3 + 3/9 + 7/27 + ...
  --------------------------------
  2S = 1 + 2/3 + 4/9 + 8/27 + ...


The expression on the right is a pure geometric series...

2S = 1/(1-(2/3)) = 1/(1/3) = 3
S = 3/2 or 1.5