To begin, note that this is working out as a sort of a pattern, where \((\frac{1}{5})^0 = 1, (\frac{1}{5})^1 = \frac15\), and then it continues on to where the exponential power keeps increasing by 1. Let's express this in sigma notation:
\(\sum _ {n=0} ^ {\infty} (\frac15)^n\), because there is no ending to the original sum, so we assume that it goes on forever to infinity. Now, we evaluate the sum, using the formula for geometeric series: \(a(\frac{1-r^n}{1-r})\), where we know a is the first term, which is 1. R is the common ratio, which we see that we are raising 1/5 to some power, so the common ratio is 1/5. Then, we have n, which is the number of terms. This is infinity, because the index of the summation ends at infinity. So, writing this out gives:
\(a(\frac{1-r^n}{1-r}) \implies 1(\frac{1-(\frac15)^{\infty}}{1-\frac15})\).
When raising something to the infinite power, we conclude that it is 0, so for the numerator, we have 1 - 0, which is 1, and for the denominator, we have 1 - 1/5, which is 4/5. 1 divided by 4/5 is 1 times 5/4, so the answer is \(\boxed{\frac54}\).
Please correct me if I made a mistake, I am always willing to learn! 
