Question 407934
This sum is represented in sigma notation by {{{64*sum((2/3)^i, i = 0, infinity)}}}.


If you know that an infinite geometric series {{{sum(ar^i, i = 0, infinity)}}} converges if and only if {{{-1 < r < 1}}}, we conclude that the series converges to {{{a/(1-r)}}}. Here, {{{a = 64}}} and {{{r = 2/3}}}, so


{{{64*sum((2/3)^i, i = 0, infinity) = 64/(1 - (2/3)) = 192}}}.