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Eddie Price Sum of a Geometric Series Spring 2018
Consider the geometric series a + ar + ar^2 + ar^3 + ... + arn^ + ... with first term a and common ratio r. We know that if โ 1 < r < 1, then this has the sum (^1) โar. In class, I gave you the correct way to see that this works, by investigating the limit of the partial sums. But there is another way to see that if the series converges, then its sum must be (^1) โar.
Let S = a + ar + ar^2 + ar^3 + ... + arn^ + .... If you assume that the sum of the series makes sense and that you can use normal algebra rules on it, then the following argument makes sense:
Multiply both sides by r:
rS = r
a + ar + ar^2 + ar^3 + ... + arn^ + ...
= ar + ar^2 + ar^3 + ar^4 + ... + arn^ + ...
Since there are infinitely many terms, we see that the only difference between this series and our original series S is a. In other words,
a + rS = S
We want to solve for S, so we subtract rS from both sides:
a = S โ rS
Factoring out S, we get
a = (1 โ r) S
Dividing both sides by 1 โ r, we get
S =
a 1 โ r
So this shows that if the series has a sum, then the sum is (^1) โar. But we canโt tell from this method when the series actually has a sum.
(For example, if a = 1 and r = 2, then the series is 1 + 2 + 4 + 8 + 16 + ..., which clearly diverges to โ, and does not have the value (^1) โar = (^1) โ^12 = โ1.)
This is why we needed to look at the limit of partial sums in class.
(See next page for an explanation on partial sums.)
Eddie Price Sum of a Geometric Series Spring 2018
In class, I told you that the nth partial sum of the geometric series a + ar + ar^2 + ar^3 + ... + arn^ + ... is a ยท 1 โr
n+ 1 โr (provided that^ r^6 = 1). I will now show you why this is true.
Recall, the nth partial sum is simply the sum of the first n terms: a+ar +ar^2 +ar^3 +...+arn. We will let Sn denote this sum. Now, we can use the same sort of argument as on the previous page! (Because Sn definitely makes sense being the sum of finitely many terms.)
Multiply both sides by r to get
rSn = r
a + ar + ar^2 + ar^3 + ... + arn
= ar + ar^2 + ar^3 + ar^4 + ... + arn^ + arn+
Now, since there are only finitely many terms here, we have a bit of a difference from before. In order to get our original sum Sn, we have to add a and subtract arn+1.
a + rSn โ arn+1^ = Sn
Since we want to solve for Sn, we subtract rSn from both sides to get
a โ arn+1^ = Sn โ rSn
We factor out a on the left hand side and S on the right hand side:
a
1 โ rn+
= (1 โ r) Sn
Now, we divide both sides by 1 โ r to get
Sn = a ยท
1 โ rn+ 1 โ r This gives the partial sum formula I described in class. And of course, analyzing the limit as n โ โ, we see that we get lim nโโ rn+1^ = 0 whenever โ 1 < r < 1, and the whole formula
diverges whenever r โค โ1 or r โฅ 1.
