What is the proof for the geometric series formula?

[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?

 

[CompuChip]

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

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

If |r| > 1, then definitely |r^k| > 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.

 

[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 license to extend this kind of algebraic manipulation to an infinite sum depends on some unstated rule. What exactly is the rule?

 

[Kyouran]

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 license to extend this kind of algebraic manipulation to an infinite sum depends on some unstated rule. What exactly is the rule?

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}

 

[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 learned 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?

 

[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}
instead of
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).

 

[Rasalhague]

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.

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?

 

[Rasalhague]

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

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)?

 

[CompuChip]

Since you insist on rigour, let us be rigorous.

Note that the summation S we are talking about is actually defined as
<br /> S \equiv \sum_{k = 0}^\infty a r^k \equiv \lim_{n \to \infty} S_n<br />
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 S_n:
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 r^k = 1 and S_n = 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.

 

[Rasalhague]

Thanks again, CompuChip. It seems much clearer now.

Since I said we would be rigorous, I will leave it to you to prove this by mathematical induction.

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.

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

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.

 

[benorin]

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

However, if \left | r \right | &gt; 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 | &lt; 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.

 

[CompuChip]

Actually, I don't really follow your induction step.
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}

 

[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.