# Convergence of geometric series with |r|>1

The geometric series $\sum_{n=0}^\infty ar^n$ with $a, r \in \mathbb{R}$ converges to $\frac{a}{1-r}$ iff $|r| < 1$.

Given this proof:

$\sum_{n=0}^\infty ar^n = a + \sum_{n=1}^\infty ar^n = a + r\sum_{n=0}^\infty ar^n$ [1],

where we can clear $\sum_{n=0}^\infty ar^n$ as

$(1-r) \sum_{n=0}^\infty ar^n = a$,

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r} \blacksquare$,

my question is: where have been used the fact that $r$ must be $|r|<1$? To me, all the steps done in [1] are true no matter how $r$ is.

• That "proof" is no proof at all - none of it makes any sense unless you already know the sum converges. – David C. Ullrich Jun 28 '18 at 15:38
• @DavidC.Ullrich why is that? – Carlos Navarro Astiasarán Jun 28 '18 at 15:41
• The proof is incomplete. It proves that when the series converges it does converge to $\frac a{1-r}$ but nothing more. It does nothing to prove that it converges if and only if $|r| < 1$. – fleablood Jun 28 '18 at 15:50
• Where did you get this bizarre and unacceptable proof? – fleablood Jun 28 '18 at 15:55
• But it's not a proof of where it converges. That is preciely what it doesn't prove. – fleablood Jun 28 '18 at 16:31

This proof is not valid since the proof already assumes convergence. The following implication makes no sense

$$\sum_{n=0}^\infty ar^n = a + r\sum_{n=0}^\infty ar^n \implies (1-r) \sum_{n=0}^\infty ar^n = a$$

unless you assume the convergence of $\sum_{n=0}^\infty ar^n$ (because assume it diverges, then $\sum_{n=0}^\infty ar^n$ is undefined. Can you do arithmetic with undefined objects?).

Here is a proof that $1=0$: $$\sum_{n=0}^\infty 1 = 1 + \sum_{n=0}^\infty 1 \implies 1=0$$

• I wouldn't say it proves "nothing". It proves if the series converges than it converges to $\frac a{1-r}$. But that was the absolute least important thing that needed to be proven. Obviously the most important thing to prove is when the series converges which this "proof" bizarrely seemed to blithely ignore. – fleablood Jun 28 '18 at 15:54
• Thank you, I will rephrase my statement. – yakobyd Jun 28 '18 at 15:56

The proof is incomplete.

To be complete it must prove.

1) the series does not converge if $|r| \ge 1$.

2) the series converges if $|r| < 1$.

3) when the series converges it converges to $\frac a{1-r}$

The proof does 3) but totally ignores the first two.

The proper proof is to show find the limit of finite sums:

For finite $n$, $\sum_{i=0}^n ar^n$ can be shown to be equal to $a\frac {r^{n+1} - 1}{r-1}$ (assuming $r \ne 1$. If $r=1$ then it is clear that $\sum ar^i = n*a$ which clearly diverges.)

(Because $(r-1)\sum\limits_{i=0}^n ar^n = \sum\limits_{i=0}^{n} (ar^{i+1} - ar^i) =\sum\limits_{i=1}^{n+1}ar^i - \sum\limits_{i=0}^{n}ar^i=ar^{n+1} - 1$.)

This is continuous so $\lim\limits_{n\to \infty} \sum_{i=0}^n ar^n=\lim\limits_{n\to \infty}a\frac {r^{n+1} - 1}{r-1}$ which converges $a\frac {K - 1}{r-1}$ if and only if $r^{n+1}$ converges to $K$ and $r \ne 1$.

It's easy to show that $r^{n+1}$ converges to $0$ if $|r| < 1$, converges (is) $1$ if $r = 1$ (which we've ruled out for other reasons), and does not converge otherwise.

So $\sum_{i=0}^\infty ar^n = \lim\limits_{n\to \infty} \sum_{i=0}^n ar^n=\lim\limits_{n\to \infty}a\frac {r^{n+1} - 1}{r-1}= a\frac {0 - 1}{r-1} = \frac a{1-r}$ if and only if $|r| < 1$.

• (1) is false unless you assume that $a \ne 0$ (which is a trivial edge case, granted, but still needs to be explicitly excluded). – Kevin Jun 29 '18 at 2:10
• Yes, I guess so. – fleablood Jun 29 '18 at 2:59

$$\sum_{n=0}^\infty ar^n = a + \sum_{n=1}^\infty ar^n = a + r\sum_{n=0}^\infty ar^n$$

As several comments and answers point out, using the $\infty$'s in the sums assumes that the limits of the partial sums exist - that's the definition of an infinite sum.

What you can say is this identity for finite sums $$\sum_{n=0}^N ar^n = a + \sum_{n=1}^N ar^n = a + r\sum_{n=0}^{N-1} ar^{n+1}$$

You can't finish your "proof" from here.

To summarize: if you somehow establish that the sum exists, then your argument correctly finds its value.

If $r\geq 1$, then the sum diverges to $\infty$, and you can't subtract $\infty$ from $\infty$. If $r \leq -1$, then the sum does not exist.

• I know that if $r>=1$ then in diverges to $\infty$, but I want to know, in such case, what is wrong with the proof given in my question. – Carlos Navarro Astiasarán Jun 28 '18 at 15:38
• @CarlosNavarroAstiasarán As I said, it fails when you are subtracting $\infty$ from $\infty$. – Botond Jun 28 '18 at 15:41
• So ... that doesn't me if $|r|<1$ that the series converges. – fleablood Jun 28 '18 at 15:44