Posted by Haile on .
Express the repeating decimal 0.513 (the 13 is repeating so the decimal is 0.5131313131313...) as a fraction in lowest terms using the infinite geometric series method.

Math 
Reiny,
0.5131313131313...
= .5 + .013131313...
= .5 + .013 + .00013 + .0000013 + ...
= .5 + an infinite geometric series
for the GS, a= .013 , r = .01
sum ∞ = a/(1r) = .013/(1.01) = (13/1000) / (99/100)
= 13/990
so sum∞ = 1/2 + 13/990 = 254/495
easier way:
for numerator, write down all digits to the end of the first repeat, from that subtract the digits that don't repeat : 513  5 = 508
for denominator, write down a 9 for each repeating digit, followed by a 0 for each nonrepeating digit : 990
so ..
.5131313 = 508/990 = 254/495