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sigma(k=n)(infinity) (1/3)^k is equal to?? (non sigma notation version, only numbers)

sigma(k=n)(infinity) (1/3)^k is equal to?? (non sigma notation version, only numbers)

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Answer #1

Clearly this is an geometric progression and the sum of this G.P is

A/1-R,

Where A is the first term and R is the common ratio of this series.

Here the first term is 1/3^n

And the common ratio is also 1/3^n

Hence the sum is

((1/3)^n)/(1-(1/3^n)) = 1/(3^n -1)

Because k=n to k= infinity

So this is the form of sum.

If k starts from 1 and end to infinity then substitute n=1 in that sum.

i.e 1/(3-1)=1/2

.

I've done for the both condition , if you haven't satisfied then ask me your doubt in comment section.

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