A geometric series is the sum of terms of the form a*(r^k), where a and r are constants and the sum is over k. The nth partial sum of the geometric series is the sum of a*r^k from k=0 to k=n. We'll call the nth partial sum S(n). So,

S(n)=a+a*r+a*r^2+…+a*r^(n-1)+a*r^n

These sorts of series are useful for analyzing many different, interesting situations. Next, I'll show how to find the value of the nth partial sum and the infinite series. I like this because it's an easy proof, but it shows a really interesting result. One may notice two different ways to get the (n+1)th partial sum (meaning the sum of the geometric series from 0 to n+1).

S(n+1)=a+a*r+…+a*r^n+a*r^(n+1)= S(n)+a*r^(n+1)

You can also get it in a less obvious way, because

S(n+1)=a+(a+a*r+…+a*r^n)*r=a+S(n)*r

Now, you can stick these two equations together by taking one and using it to substitute in for S(n+1) in the other:

S(n)+a*r^(n+1)=a+S(n)*r

Solving for S(n) we get S(n)=a*(1-r^(n+1))/(1-r) to be the value of the nth partial sum, the value of the geometric series from k=0 to k=n. The best thing about this is that to get the value of the infinite series, we just have to take the limit of S(n) as n goes to infinity, provided it exists. The limit exists when |r| < 1. In that case, as n goes to infinity, r^(n+1) goes to zero (because every time you multiply two numbers with |r| < 1 together you get an even smaller one). Thus, for n approaching infinity,

S=a(1-0)/(1-r)=a/(1-r)