Problem: Let sβ be the limiting sum of the geometric series 4β38β+916βββ¦, as the number of terms increases without bound. Then s equals:
Answer Choices:
A. a number between 0 and 1
B. 2.4
C. 2.5
D. 3.6
E. 12
Solution:
Since 1+a+a2+β¦=1βa1β when β£aβ£<1, the required limiting sum equals 1β(β32β)4β=2.4