why infinite geometric series sum doesn't work for |ratio|>1 Actually, I has asked this question about 7 years ago in stackoverflow, it got closed due not being programming related and was suggested to ask in math stackexchange. I am asking again, as I could not get a satisfactory answer.
Recently I was calculating a geometric series sum and then a doubt occurred about why the sum doesn't exist for r>1.
Say, S is a geometric series sum.
S  = 1 + r + r^2 + r^3 + r^4 + ...
rS =     r + r^2 + r^3 + r^4 + ...
-----------------------------------
S(1-r) = 1
S = 1/(1-r)

Here, I never assumed anything about the value of r. But then why it is valid for |r|<1 and not for |r|>=1. I know, if it is valid, the sum of infinite positives numbers will be negative and will open a whole can of worms. But, WHY? I couldn't think of any sound logic to it.
Edit:
I know and have seen this derivation, first for finite series then extending it for infinite. And then, it is fine.
But, then recently I was reading about infinity and its cardinality. In that there was a set of natural numbers A, another set of natural number B without 1, but then it says that still there is a 1-1 mapping between both set,  x in A mapped with x+1 in B. There is no element in A which is not mapped into B and vice-versa. So, the cardinality of both set is same.
So, from the same logic, if I don't go by finite and then extending it to infinite. I just start with infinite series, then every element in rS is in S with opposite sign. There is no element which is not mapped except the initial 1. And that was where my logic failed.
 A: Essentially it is a $\infty - \infty$ problem: when $S=\infty$ and $rS=\infty$, $S-rS$ is $\infty-\infty$ which is an "indeterminate form". In general you can't add two extended real numbers unless either one of them is finite or they have the same sign.
The way to avoid this problem is to look at the finite sum, in which case the same procedure yields
$$(1-r)S=1 - r^{n+1}$$
and now you cannot send $n \to \infty$ and get a finite limit on both sides unless $|r|<1$ or $r=1$. (And in the case $r=1$ you still can't evaluate the infinite series this way: why?).
In view of your last paragraph, I should point out that the same issue occurs with cardinalities. For finite sets, with $B \subset A$, $|A \setminus B|=|A|-|B|$. But with infinite sets, you cannot generally determine the cardinality of $A \setminus B$ only knowing $|A|$ and $|B|$. For example $|\mathbb{N} \setminus \{ 2,3,4,\dots \}|=1$ but $|\mathbb{N} \setminus \{ 2,4,6,\dots \}|=|\mathbb{N}|$.
A: Unless $a_n \to 0$ we can't have $\sum_{k=0}^{\infty} a_n$ converge so $S = 1 + r+ r^2 + r^3+....$ is not possible as $r^n \not \to 0$.
For $\sum r^k$ this should be very clear if we consider that $\sum r^k = L$ means that for any $\epsilon > 0$ then there is an $N\in \mathbb N$ so then for all $n> N$ we have $|L - \sum_{k=0}^n r^k|<\epsilon$.
Let $\epsilon =\frac 12$ and $|r| \ge 1$ so $|r^m|\ge 1$ for all $m$.  So if $|L-\sum_{k=0}^m r^k| < \epsilon$ then $|L - \sum_{k=0}^{m+1}r^k|= |(L-\sum_{k=0}^m r^k)- r^{m+1}|< \epsilon$ would imply that $|r^{m+1}|=|(L - \sum_{k=0}^{m+1}r^k)- (L - \sum_{k=0}^{m+1}r^k)|\le |L - \sum_{k=0}^{m+1}r^k|+|L - \sum_{k=0}^{m+1}r^k| < 2\epsilon < 1$.  A contradiction.
So there can not be any $N$ so that if $m$ and $m+1> N$ then $|L - \sum_{k=0}^{m+1}r^k|<\frac 12$ and $|L - \sum_{k=0}^{m+1}r^k|<\frac 12$ for any $L$.  As that is impossible, claiming $\sum_{k=0}^{\infty} r^k = L$ for any $L$ is meaningless.
