Convergence of geometric series with |r|>1

Question

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

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

leading to

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

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.

Answer 1

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 $1=0$: $$ \sum_{n=0}^\infty 1 = 1 + \sum_{n=0}^\infty 1 \implies 1=0$$

Answer 2

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.

Answer 3

Your problem is here:

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

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.

Answer 4

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 $|r| < 1$.