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}