SOLUTION: Which of the following series is convergent? 1 + 1/2^2 + 1/2^4 + 1/2^6 + ... 2 + 4/2^2 + 8/3^2 + 16/4^2 + ... 1/2+ 2/3 + 3/4 + 4/5 + ... 1 + (3/2)^2 + (3/2)^4 + (3/2)^6

Algebra ->  Finance -> SOLUTION: Which of the following series is convergent? 1 + 1/2^2 + 1/2^4 + 1/2^6 + ... 2 + 4/2^2 + 8/3^2 + 16/4^2 + ... 1/2+ 2/3 + 3/4 + 4/5 + ... 1 + (3/2)^2 + (3/2)^4 + (3/2)^6      Log On


   

Question 1035615: Which of the following series is convergent?
1 + 1/2^2 + 1/2^4 + 1/2^6 + ...
2 + 4/2^2 + 8/3^2 + 16/4^2 + ...
1/2+ 2/3 + 3/4 + 4/5 + ...
1 + (3/2)^2 + (3/2)^4 + (3/2)^6 + ...
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Since the first one is the only one with decreasing terms...r = 1/4...it converges to
S%5Bn%5D+=+a%5B1%5D%2F%281-r%29+=+1%2F%281-%281%2F4%29%29+=+1+%2F+%283%2F4%29+=+4%2F3