Problem in sum of infinite geometric series? Let $s=1+r+r^2+r^3+\ldots$
Then $rs=r+r^2+r^3+\ldots$
Then $s-rs=1$, so $s(1-r)=1$, and thus $s=\dfrac{1}{1-r}$
This is a taken from Wikipedia's article on geometric series. I think there is a problem with this. We are assuming that the sum is $s$ a number and then doing operations on it. Well what if $s$ is not a number, infinity perhaps? We have not assumed or proved convergence of the geometric series. Thus I think there is a hole in this proof. Am I right or wrong? 
If there is that gap we must prove the convergence. I do not see any method using limits or anything else. Can you help me?  
 A: Wikipedia starts the explanation proving $$s = 1 + r + ...+r^{n-1} = \frac{1-r^n}{1-r}$$. 
This is still a number, since the sum is finite. Now take the limit $n$ to $\infty$ and you have your solution whenever $|r|<1$. The next part was just a review of what was done before.
EDIT: Now to prove that $\lim_{n\rightarrow\infty} r^n=0$ if $|r|<1$:
One general idea would be that whenever $|r|<1$, the infinite sequence
$$r,r^2,r^3,...,r^n,...$$
is monotonically decreasing and bounded by $0$. Therefore, it has a limit. 
To prove that this limit is $0$ is the same as to prove that for every real $\epsilon>0$, there is an $n_o\in \mathbb{N}$ such that $|r^{n_o}|<ϵ$. This is equivalent to saying $$|r^{n_o}|<ϵ$$ 
Now let's make $\frac{1}{r}=r_{inv}$. Therefore $|r_{inv}|>1$, and the infinite sequence $$r_{inv},r_{inv}^2,...,r_{inv}^n,...$$
is monotonically increasing and unbounded. This is the same as saying that for every $M\in\mathbb{R}$, there is an $n_1 \in \mathbb{N}$ such that $|r_{inv}^{n_1}|>M$. But this applies for every real M, so we pick one such that $M>\frac{1}{\epsilon}$, giving $$|r_{inv}^{n_1}|>M>\frac{1}{\epsilon}$$ Inverting the last equation, we get $$|r^{n_1}|<ϵ$$
A: You're correct. The assumption in this argument is that the sum of the series exists, i.e., that the partial sums form a convergent sequence. You can, however, read the argument as follows:
"If the sum $s$ exists and is finite, then the following algebraic steps show that it must in fact be $\dfrac{1}{1-r}$. When $|r| > 1$, it's clear that the sum diverges. All that remains is to show that for $|r| < 1$, the series converges. "
Then there's some work to do. One possibility is to show that for $|r| < 1$, the Cauchy property holds, in which case the series converges. 
A: Of course convergence must be proven, and it is, for $|x|<1$.
There is a closed formula for finite geometric series:
$$\sum_{k=0}^naq^k=\frac{a(q^n-1)}{q-1}$$
Limiting n to infinity gives the formula for infinite sum.
EDIT:
It remains to show that for any $q$ for which $|q|<1$, $\lim_{n\to\infty}q^n=0$
Put $r=\frac1q$. Then $|r|>1$ and $\lim_{n\to\infty}r^n=\infty$ and $\lim_{n\to\infty}q^n=\lim_{n\to\infty}\frac1{r^n}=\frac{1}{\infty}=0$
It is not exactly formal and the case $q<0$ (In this case the absolute value need to be considered) but it is a form for proof.
Here a proof that $\lim_{n \to \infty} n x^{n} = 0$, I think it imlies.
A: I think convergence of the geometric series is required (i.e., $|r|<1$) when we do operations on it. Otherwise we may get unpleasant case such as $\infty-\infty$.
