How do you find the sum of the infinite geometric series 4/5+4/15+4/45...?

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Answer 1 · 2018-07-28T08:59:29

The sum is #=6/5#

Let general term of the GP be

#u_n=ar^n#

The sum of an infinite GP is

#S_oo=sum_(k=0)^ooar^k#

The sum is

#S_oo=a/(1-r)# if the common ration is #|r|<1#

Here,

The first term is #a=4/5#

And

The common ration is #r=1/3#

Therefore,

#S_oo=(4/5)/(1-1/3)=4/5*3/2=6/5#