Where have I erred in proving $\sum_{k=0}^\infty ar^k = {a\over 1-r}$? I'm familiar with the infinite geometric series, its convergence conditions, and the formula for the value of convergence:
$$S_{\infty} = {a\over 1-r}\quad |r| < 1$$
...for an infinite geometric series with start term $a$ and common ratio $r$.
I've seen this proven a few different ways, and recently devised my own simple proof founded in simple algebraic manipulations. It is intuitively clear to me that $|r|<1$ is a necessary condition for convergence. However, in the below proof, I seem to find that the value of convergence is $a/(1-r)$ without requiring $|r|<1$ (only $r\neq 1$ seems to be necessary).
Have I erred somewhere? Where should the convergence condition come into play?

Proof.
Consider the sum: \begin{align} \sum_{k=0}^\infty ar^k  &= ar^0 + ar^1
 + ar^2 + \cdots \\ & = a + ar + ar^2 + \cdots \\ & = a(1 + r + r^2 + \cdots) \end{align}
Now, consider the product:
$$(1-r)(1 + r + r^2 + \cdots)$$
\begin{align} \require{cancel} & = (1 + \cancel{r} + \cancel{r^2} +
 \cdots) + (\cancel{-r} - \cancel{r^2} - \cdots) \\ & = 1 \end{align}
Hence:
$${1\over 1-r} = (1+r+r^2+\cdots)\quad r\neq 1$$
Therefore:
$${a\over 1-r} = a(1 + r + r^2+\cdots) = \sum_{k=0}^\infty ar^k\quad r\neq 1$$

 A: What you showed is that "if the sum $\sum_{k=0}^{\infty}ar^k$ converges and $r\neq 1$ then $\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}$". And you certainly gave a correct proof of this fact, because by assuming the series converges, all the manipulations can be justified.
However, you haven't answered the question of when the sum $\sum_{k=0}^{\infty}ar^k$ converges. So, just emphasizing again: you gave a correct proof, but it's a proof of a completely different question. You haven't answered the question "for what values of $r$ does the sum $\sum_{k=0}^{\infty}ar^k$ converge"?
The full answer is

*

*If $a=0$, the sum clearly converges for all values of $r$.

*If $a\neq 0$, the sum converges if and only if $|r|<1$.

The proof of this relies on the observation that for $r\neq 1$, $\sum_{k=0}^nr^k=\frac{1-r^{n+1}}{1-r}$. The limit of this as $n\to\infty$ exists in $\Bbb{R}$ if and only if $|r|<1$. The advantage of this approach is that we're literally using the definition of a series as a limit of partial sums, so it makes it very obvious where the limiting condition occurs and what restrictions have to be placed on $r$.
