An infinite geometric series has 1 and 1/5 as its frist two terms: 1, 1/5' 1/25' 1/125' what is the sum, s,of the infinite series?
a 1/4
b1/25
c 5/4
d 1

 you have a geometric series 

since you are summing the powers of 1/5, this converges 
(1/5 < 1) 

it's equal to 1/(1-1/5) 
=1/(4/5) = 5/4 

for geometric series, the sum is always 
x = 1/(1-r) 
where r is the ratio of successive terms. 

if you set the sum equal to x: 
x = 1+r+r^2+r^3... 
and multiply each term by r 
rx= r+r^2+r^3+ 
then subtract 

x-rx = 1+r-r+r^2-r^2+r^3-r^3+ 

x-rx = 1 
x(1-r)=1 
x=1/(1-r)