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