Evaluate the infinite geometric series: 13+16+112+124…
First term is 1/3.
Common ratio is (1/6)/(1/3) = 1/2.
Using the formula sum = a/(1 - r), sum = 131−12=23
Thanks so much!
@MaxWong and @Anthrax
Observe that we can say
13+16+112+⋯=13(1+12+14+⋯)=13∞∑n=0(12)n.
Given this, our common ratio r is r=12. Then, by the well-known infinite geometric series formula, we have
13∞∑n=0(12)n=13⋅11−12=23.