TITLE: Is this cancellation argument used to find the sum of a geometric series correct?
QUESTION [0 upvotes]: Here is a common argument used to prove that the sum of an infinite geometric series is $\frac{a}{1-r}$ (where $a$ is the first term and $r$ is the common ratio):
\begin{align}
S &= a+ar+ar^2+ar^3+\cdots \\
rS &= ar+ar^2+ar^3+ar^4+\cdots \\
S-rS &= a \\
S(1-r) &= a \\
S &= \frac{a}{1-r} \, .
\end{align}
I am sceptical about the validity of this argument. It feels like there is something amiss about a proof involving infinite series that makes no mention of the fact that they are typically defined as the limit of their partial sums. The third line involves 'cancelling' all of the terms other than $a$. This makes it seem like an infinite series is actually an infinite string of symbols, rather than a limiting expression. If $a+ar+ar^2+ar^3+\cdots$ is simply a shorthand for
$$
\lim_{n \to \infty}\sum_{k=0}^{n}ar^k \, ,
$$
and $ar+ar^2+ar^3+ar^4+\cdots$ a shorthand for
$$
\lim_{n \to \infty}\sum_{k=1}^{n}ar^k \, ,
$$
then I am struggling to see how the terms actually cancel. Perhaps the third line can be written more formally as
\begin{align}
S-rS &= (a+ar+ar^2+ar^3+\cdots)-(ar+ar^2+ar^3+ar^4+\cdots) \\
&= \lim_{n \to \infty}\sum_{k=0}^{n}ar^k - \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \\
&= \lim_{n \to \infty}\left(\sum_{k=0}^{n}ar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}\left(a +\sum_{k=1}^nar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}a \\
&= a \, .
\end{align}
There is another concern I have. Geometric series only converge when $|r|<1$. However, the argument used above seems to apply regardless of whether $|r|<1$, which would yield nonsensical results such as
$$
1+2+4+8+\cdots = -1 \, .
$$
So is the argument rigorous, and if so, why are my fears misplaced?

REPLY [2 votes]: Let
$$S_n=\sum_{k=0}^n ar^n.$$
Then
$$(1-r)S_n=S_n-rS_n=a-ar^{n+1}$$ and
$$S_n=a\frac{1-r^{n+1}}{1-r}.$$
This is absolutely rigorous for any $r\ne1$.
And from this, for $|r|<1$,
$$S=\lim_{n\to\infty}S_n=\frac a{1-r}.$$
Now, for $|r|\ge1$, the infinite sum diverges as does the limit of $r^{n+1}$, so there is no paradox.
{"set_name": "stack_exchange", "score": 0, "question_id": 3953674}