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,

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