Prove that :

(9^1/3 . 9^1/9 . 9^1/27 ….∞) = 3.

Using the properties of exponents:

The above term can be written as

Let S = …(1)

We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression.

Let m =

Common ratio = r =

Sum of infinite GP = ,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 = and r =

⇒ m =

From equation 1 we have,

S = 9^m = 9^1/2 = 3 = RHS

Hence Proved