Find the values of x for which the series 1/x + 1/x^2  + 1/x^3+…  converges and express the sum as a function of x

\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+...

Solution

The geometric sum formula for infinite terms is given as:

If  common ration |r| < 1, S_\infty=\frac{a}{1-r}

If |r| > 1, the series does not converge and it has no sum.

Where

a is the first term of series

r is the common ratio of series

n is the number of terms of series

This is a geometric series with ratio 1/x. It converges for |1/x|<1  that is , for |x|>1 The sum is

\frac{\frac{1}{x}}{1-\frac{1}{x}}=\frac{1}{x-1}