Let the first term of a GP series is "a" and the common difference is "d" then sum of n terms of an infinite series $$ S_n = \frac{a (1 - d^n)}{1 - d} $$ $$ S_n = \frac{a - ad^n}{1 - d} $$ $$ S_n = \frac{a}{1 - d} - \frac{ad^n}{1 - d} $$
Case(1): if \(d \lt 1\) then the value of \(ad^n\) tends to zero, hence the sum of infinite GP series $$ S_n = \frac{a}{1 - d} $$
Case(2): if \(d \gt 1\) then the sum of infinite GP series will also be infinite.
Example(1): Find the sum of the given infinite GP series? $$ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.... $$
Sulution: From the given series
first term of the series (a) = 1
common difference (d) = \(\frac{1}{2}\)
and n is infinite then $$ S_n = \frac{a}{1 - d} $$ $$ S_{\infty} = \frac{1}{1 - \frac{1}{2}} = 2 $$ Hence the sum of the given infinite series is 2.
Example(2): Find the sum of the given infinite GP series? $$ \sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + \frac{1}{4\sqrt{2}} +.... $$
Sulution: From the given series
first term of the series (a) = \(\sqrt{2}\)
common difference (d) = \(\frac{1}{2}\)
and n is infinite then $$ S_n = \frac{a}{1 - d} $$ $$ S_{\infty} = \frac{\sqrt{2}}{1 - \frac{1}{2}} $$ $$ S_{\infty} = \frac{\sqrt{2}}{\frac{1}{2}} $$ $$ S_{\infty} = 2 \sqrt{2} $$ Hence the sum of the given infinite series is \(2 \sqrt{2}\).