Question

Find the sum of the infinite series 1/3+1/9+1/27 if it exists.

Answer 1

infinite? do you mean 1/3+1/9+1/27+1/81+...?

Answer 2

@sara_singh12 can you tell me that is it GP or AP ?

Answer 3

@Clouds yes I think it is \[\large{\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + ...}\]

Answer 4

in fact,\[p+p^2+p^3+...+p^n=\frac{ p-p^{n+1} }{ 1-p }\]

Answer 5

I think @sara_singh12 is not present here , though I am putting the formula up here so that will help her to solve the problem :

\[\large{S _ \infty = \frac{a}{1-r}}\]

Answer 6

it is an infinite gp with common ratio of 1/3
so, sum = a/(1 - r)
= 1/3 divided by ( 1 - 1/3)
= 1/2

Answer 7

More explanation

First term = \(\frac{1}{3}\)

Common ratio = \(\large{\frac{a_2}{a_1}}\) = \(\large{\frac{\frac{1}{9}}{\frac{1}{3}}}\) = \(\large{\frac{1}{3}}\)

Therefore : \[\large{S_ \infty = \frac{\frac{1}{3}}{1- \frac{1}{3}}}\]
\[\large{S_\infty = \frac{\frac{1}{3}}{\frac{2}{3}}}\]
\[\large{S_\infty = \frac{1}{2}}\]