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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! 

Good Job you are correct

THX, mobro....here's a slightly easier way

coomon ratio   = 1/5          a  = first term   = 1

Sum    =    a / (1 - r)  =  1  /( 1 -1/5)  =  1 / (4/5)  = 5/4