Geometric series formula

1. Jan 28, 2010

Rasalhague

$$\sum_{k=0}^{\infty} ar^k = \frac{a}{1-r}$$

This equation isn't valid, for real numbers, unless $\left | r \right | \leq 1$. I can see that if $r = 1$ the denominator is be zero, but what about the other cases? The derivation I've seen is

$$\sum_{k=0}^{\infty} ar^k = \sum_{k=0}^{\infty} ar^k \cdot \frac{1-r}{1-r} = \frac{a}{1-r}.$$

What part of this derivation depends on the assumption that $\left | r \right | \leq 1$?

2. Jan 28, 2010

CompuChip

The part where you assume that the (formal) sum on the left hand side converges.

If |r| > 1, then definitely |rk| > 1 for all integer k in the summation, and in fact, $|r^k| \to \infty$ as $k \to \infty$.

Thus,
$$\left| \sum_{k \ge 1} r^k \right| = \infty$$.

3. Jan 28, 2010

Rasalhague

Thanks, CompuChip. Let's assume that $|r| > 1$. The derivation goes:

$$\sum_{k=0}^{\infty} ar^k = \sum_{k=0}^{\infty} ar^k \cdot \frac{1-r}{1-r} = \frac{a + \left ( r^1 - r^1 \right ) + \left ( r^2 - r^2 \right ) + ...}{1-r}.$$

Each $r^k \in \mathbb{R}$, and $\forall x \in \mathbb{R} : x-x = 0$, therefore $r^k - r^k = 0$. And $0 + 0 = 0$, therefore

$$\sum_{k=0}^{\infty} ar^k = \sum_{k=0}^{\infty} ar^k \cdot \frac{1-r}{1-r} = \frac{a + \left ( r^1 - r^1 \right ) + \left ( r^2 - r^2 \right ) + ...}{1-r} = \frac{a}{1-r}$$

The reasoning doesn't explicitly restrict the absolute value of $r$ to being less than one, so I'm thinking perhaps the licence to extend this kind of algebraic manipulation to an infinite sum depends on some unstated rule. What exactly is the rule?

4. Jan 28, 2010

Kyouran

Actually when k goes to infinity, the sum goes to

$$\frac{a-ar^{\infty}}{1-r}$$

Only if you assume that |r| is smaller than or equal to 1 can you say that it equals to

$$\frac{a}{1-r}$$

Last edited: Jan 28, 2010
5. Jan 28, 2010

Rasalhague

Could you elaborate a bit on the meaning of

$$\frac{a-ar^\infty}{1-r}?$$

What are the formal rules for manipulating expressions like $x^\infty$, and how did you derive this formula? Is this notation standard? In talking about the field of the real numbers with addition and multiplication as usually defined, the only kind of exponents that I've learnt about are real numbers. Did you go through a step like

$$\frac{a+ar^{\infty_1}-ar^{\infty_2}}{1-r}$$

with two different infinities, one of which somehow outweighs the other?

6. Jan 28, 2010

Kyouran

I wasn't being formal at all when I posted that, nor are the exponents real numbers. In fact, they're integers, it is a discrete summation after all.

And this might not be the correct notation either, in the field of the real numbers, we would use a limit, but there's no such thing when talking about integers. Hence, I decided to merely write it in a manner that would make it easier to understand, but it seems it only caused more confusion.

What I meant is the following:
For each k, you have a term $$-ar^kr$$ in the nominator. It's true that this is being neutralized by the term $$ar^{k+1}$$, but I think the problem lies in stopping the summation there, not accounting for the next term, $$-ar^{k+1}r$$. In your proof, the last term is:

$$ar^k \frac{1}{1-r}$$
$$ar^k \frac{1-r}{1-r}$$

And by doing so you assume that |r| is equal to or smaller then 1

edit: the term you're forgetting is the term that determines whether it converges or diverges. By not taking into account this term you're actually assuming it converges (unless |r| = 1, in that case it diverges as well).

Last edited: Jan 28, 2010
7. Jan 28, 2010

Rasalhague

Isn't every "next term" of the form $-ar^{k+1}r$ already expressed as $-ar^{k}r$, for the next value of k in the sequence, given that k is summed over to infinity?

8. Jan 28, 2010

Rasalhague

Do you mean the rule is that we can only perform these conventional algebraic operations, multiplication or division by a polynomial, on

$$\sum_{k=1}^{\infty} ar^k$$

if the value of the sum is finite, and that they're undefined when the value of the sum is not finite (and that's why we can get nonsensical results if we apply the formula to a series that doesn't have a finite value)?

9. Jan 29, 2010

CompuChip

Since you insist on rigour, let us be rigorous.

Note that the summation S we are talking about is actually defined as
$$S \equiv \sum_{k = 0}^\infty a r^k \equiv \lim_{n \to \infty} S_n$$
where
$$S_n \equiv \sum_{k = 0}^n a r^k$$

Without loss of generality, let us set a = 1 (it's just a post-multiplication by a real number). Now work out Sn:
$$S_n = \sum r^k = \sum r^k \cdot \frac{1 - r}{1 - r} = \frac{1}{1 - r} \sum (r^k - r^{k + 1}).$$
(I'm leaving out the indices on the sum, they all run from k = 0 to n), provided r is not equal to 1 (if it is equal to 1, then rk = 1 and Sn = n + 1).

If you write this out a bit, you will see that adjacent terms cancel, and you are always left with
$$S_n = \frac{1}{1 - r} \left( r^0 - r^{n + 1} \right) = \frac{1 - r^{n + 1}}{1 - r}.$$
Since I said we would be rigorous, I will leave it to you to prove this by mathematical induction.

Now, again separating the case r = 1 (giving $S = \lim_{n \to \infty} (n + 1) = \infty$),
$$S = \lim_{n \to \infty} \frac{1 - r^{n + 1}}{1 - r} = \frac{1}{1 - r} \left( 1 - \lim_{n \to \infty} r^{n + 1} \right)$$

Of course, the limit is equal to 0 when |r| < 1, and when |r| > 1, rn+1 diverges and so does S.

Afterthought: are you familiar with real analysis? Then you might want to check out the concept of radius of convergence.

10. Jan 29, 2010

Rasalhague

Thanks again, CompuChip. It seems much clearer now.

If n = 0, then

$$\sum_{k=0}^{n}\left(r^k - r^{k+1}\right) = 1-r^1 = 1-r^{n+1}.$$

Otherwise,

$$\sum_{k=0}^{n}\left (r^k - r^{k+1}\right) = r^0-r^{n+1} + \sum_{k=1}^{n}\left ( r^k - r^k \right )$$

$$= 1-r^{n+1}$$

because

$$\sum_{k=1}^{n}\left ( r^k - r^k \right ) = 0$$

because $0+0 = 0$, so $0+0+...+0=0$.

I actually came across this in Roger Penrose's The Road to Reality where he introduces the idea of radius of convergence, so that's definitely my next port of call.

11. Jan 29, 2010

benorin

Typically, one uses the geometric series formula to expand $\frac{a}{1-r}=\sum_{k=0}^{\infty}ar^{k}$ for $\left | r \right | < 1$.

However, if $\left | r \right | > 1$, then one may use,

$$\frac{a}{1-r}=-\frac{1}{r}\cdot\frac{a}{1-r^{-1}}=-\frac{1}{r}\sum_{k=0}^{\infty}a\left( r^{-1}\right) ^{k}=-\sum_{k=0}^{\infty}a r^{-(k+1)}.$$​

BTW, a related identity is the geometric product: for $\left | x \right | < 1$,

$$\frac{1}{1-x}=\prod_{k=0}^{\infty}\left(1+ x^{2^k}\right)$$​

where the formula may be proven by either clever algebraic manipulation or combinatorial consideration of binary numbers.

12. Jan 30, 2010

CompuChip

I would expect that you write

$$\sum_{k = 0}^{n + 1} (r^k - r^{k + 1}) = \left( \sum_{k = 0}^n (r^k - r^{k + 1}) \right) + (r^{n + 1} - r^{(n + 1) + 1})$$
and then use the induction hypothesis,
$$\left( \sum_{k = 0}^n (r^k - r^{k + 1}) \right) = 1 - r^{n + 1}$$

13. Jan 30, 2010

Rasalhague

I suppose I should have stated that

$$\sum_{k=0}^{n}\left ( r^k - r^{k+1} \right )= 1 - r^{n+1}+\sum_{k=1}^{n}\left ( r^k - r^k \right ) = 1-r^{n+1}$$

is true in the case of n = 1, where

$$r^0 - r^2 + r^1 - r^1 = 1 - r^{n+1}$$

and called that the base case.

For the inductive step, it will be true that, for $n = n_0 + 1$,

$$\sum_{k=0}^{n}\left ( r^k - r^{k+1} \right ) = 1 - r^{n+1}$$

because

$$0+0=0$$

implies

$$\sum_{k=1}^{n_0+1}\left ( r^k - r^k \right ) = 0.$$