Question 393636
The ratio between two successive terms is -1/2 which is between -1 and 1 so the series converges.


We have the sum {{{sum(64*(-1/2)^r, r = 0, infinity)}}}. Factoring 64 out, this is equivalent to


{{{64sum((-1/2)^r, r = 0, infinity)}}}


The sum of all the terms of a convergent geometric sequence 1 + r + r^2 + ... = 1/(1-r), so replacing r with -1/2,


{{{64sum((-1/2)^r, r = 0, infinity) = 64*(1/(1-(-1/2))) = 64*(2/3) = 128/3}}}


Note that the words "series" and "sequence" have different meanings. {64, -32, 16, -8, ...} is a sequence. The sum of the terms in a sequence is defined as a series.