# Sum of infinite series

#### Sum of infinite GP series:

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 $$2 \sqrt{2}$$.