Problem: For the infinite series 1−1/2−1/4+1/8−1/16−1/32+1/64−1/128−… let S be the (limiting) sum. Then S equals:
Answer Choices:
A. 0
B. 2/7
C. 6/7
D. 9/32
E. 27/32
Solution:
Combine the terms in threes, as follows,
41​+321​+2561​,⋯∴ S=1−81​41​​=72​
or
Since there is absolute convergence, the terms may be re-arranged to yield the three series
1+81​+641​+…,−21​−161​−1281​−…
−41​−321​−2561​−…
S1​=1−81​1​=78​S2​=1−81​−21​​=−74​
S3​=1−81​−41​​=−72​∴S=72​