How do I find the sum of the infinite geometric series #2/3#, #- 4/9#, ...?

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Answer 1 · 2014-10-17T21:54:11

Recall: Geometric Series

#a+ar+ar^2+ar^3+cdots=a/{1-r}# if #|r|<1#.

Let us look at the posted geometric series,

#S=2/3-4/9+8/27-cdots#

by rewriting a bit to fit the form of geometric series,

#=2/3+2/3(-2/3)+2/3(-2/3)^2+cdots#

since #a=2/3# and #r=-2/3#,

#={2/3}/{1-(-2/3)}=2/5#

I hope that this was helpful.