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=1−8141=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=1−811=78S2=1−81−21=−74
S3=1−81−41=−72∴S=72