Find the sum of the following series to infinity : 1 - 1/3 + 1/3^2 -1/3^3 + 1/3^4 + ........∞

Question

Find the sum of the following series to infinity :

1- \(\frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} + .....\infty\)

Answer 1

We observe that the above progression possess a common ratio. So it is a geometric progression. 

Common ratio = r = \(\cfrac{\frac{-1}{3}}{1}\) = \(\frac{-1}{3}\)

Sum of infinite GP = \(\frac{a}{1-r}\),where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = 1 and r = \(\frac{-1}{3}\)

⇒ sum = \(\cfrac{1}{1-(\frac{1}{3})}\) = \(\cfrac{3}{4}\)