Prove that :

(91/3 . 91/9 . 91/27 ….∞) = 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 = 9m = 91/2 = 3 = RHS


Hence Proved


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