How do I find the sum of an infinite geometric series, if one exists? What's the equation?
You can make up yourself an infinite geometric series. Just pick any number that you want as your 1st. term and then multiply it by ratio greater than 1 and it will diverge...........etc.
Example: Your 1st. term=1 and the ratio is 2, then you would have:
1, 2, 4, 8, 16, 32, 64......... and so on for as long as you want. So, by definition, you cannot sum up an infinite geometric that diverges. But, you can sum up an infinite series that converges such as this one: 1 + 1/2 + 1/4 + 1/8 + 1/16..................forever =2!!.
F x {[1 - r^n] / [1 - r]}=Sum of a geometric series, where F=1st. term, r=ratio, n=number of terms.
Using this formula, you can sum up, say, the 1st. 100 terms of the above series: 1, 2, 4, 8.......etc. and you will get=1.27 x 10^30.
A G.P. has the form a + ar +a(r^2)....... + a(r^n-1) where a is the first term and r is the common ratio. General formula for sum to n terms is Sn = a{ (r^n) -1 } / (r-1)
The general formula for the sum to n terms is found like this:
Write the sum as Sn so that
Sn = a +ar + a(r^2).......... a(r^n-1)
Now multiply both sides by r to get
r Sn = ar +a(r^2) +a(r^3) .....+a(r^n-1) + a(r^n)
Now subtract Sn from rSn to get
rSn-Sn = a(r^n) - a
(If you write out rSn underneath Sn you can see how the terms cancel out to leave a(r^n)-a )
From rSn-Sn= a(r^n)-a, factor out Sn from the left hand side,and factor out a from the right hand side to get
Sn(r-1) =a{ ( r^n) - 1} so the sum of n terms is
Sn = a{ (r^n)-1} /(r-1). Note that this is not valid for r=1.
