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when we say $1+r+r^2+r^3+\cdots=S$ do we mean $S$ is the partial sum of first $n$ terms of this geometric series where $n$ goes to infinity (which is not boundless due to the existence of $n$th term) or do we say $S$ is sth that has literally infinite, boundless terms so that $S-1=rS$?

In other words, are we saying when we talk about $S$, are we talking about the partial sum of the first n terms where n is getting really really large or do we literally mean the bound does not exist at all, so that $S-1=rS$?

This matters a lot to me because it seems like there can be some values of things like that even if the partial sum is divergent!

If the meaning of $S$ is the 'limit' of partial sum then it's obvious that $S-1=rS$ if and only if the absolute value of $r$ is less than one. but when we treat $S$ as sth that does not have the 'nth term (the bound)' no matter how far away then I can't think of any reason of saying $S-1=rS$ is wrong. So, to sum up, is $S$ limit of the partial sum or is it sth that is endless? I hope I can get satisfying answer of this question.

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We say $S$ is the limit of the sequence of partial sums (under the usual assumption of $|r| < 1$). – J. M. May 14 '11 at 5:27
and why not the other one? is it because it's non sense? or are there some other reasons? – Jee Hoon Park May 14 '11 at 5:29
Or, to put it another way: given some tiny positive number $\varepsilon$, we can always find an integer $n$ such that the absolute difference between $S$ and the $n$-th partial sum is less than or equal to $\varepsilon$... that's another way of thinking about the limit concept. – J. M. May 14 '11 at 5:30
See also this general question about geometric series: . – Arturo Magidin May 14 '11 at 5:35
@Jae Hoon: Note also that if you let $S_n$ be the partial sum, $$S_n = 1+r+r^2+\cdots+r^n$, then $S_n$ itself satisfies $S_n - 1 = rS_{n-1}$, or $S_n - rS_n = 1-r^{n+1}$. From this you can get the value of $S_n$ in terms of $r$, and then determine whether $\lim_{n\to\infty}S_n$ exists and what its value is if it does. – Arturo Magidin May 14 '11 at 5:46
up vote 8 down vote accepted

The official meaning of the assertion that $$1+r+r^2+r^3+\cdots=S$$ is that $$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n) =S$$.

By the usual formula for the sum of a finite geometric series, $$1+r+r^2+\cdots +r^n=\frac{1-r^{n+1}}{1-r}$$ (unless, of course, $r=1$). It is reasonably clear that if $|r|<1$, $$\lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}$$ and that if $|r|\ge 1$, then $$\lim_{n \to \infty}(1+r+r^2+\cdots +r^n)$$ does not exist.

But that is not really your question. What I think you are saying is roughly this: let $$S=1+r+r^2+r^3+\cdots$$ Then at the level of formal "algebraic" manipulation, we have (??) $$S-1=rS$$ and "therefore" (??) $S=1/(1-r)$ (at least if $r \ne 1$).

But do "formal" manipulations always yield correct results? Not necessarily. For example, you have probably already seen plausible-looking algebraic manipulations that appear to yield the absurd "result" that $0=1$ (usually here the flaw is a carefully hidden division by $0$.)

The formal manipulation yields, if we take $r=1/2$, the perfectly correct result that $$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots =2$$ This makes physical sense. If you look at the real number line, and put a dot at $1$, then at $1+1/2$, then at $1+1/2+1/4$, and so on, your dots are clearly approaching $2$, and are after a while indistinguishably close to $2$.

Now look at the formal sum $$S=1+2+2^2+2^3+\cdots$$ Purely formal manipulation then yields that $S=-1$. Does this make any kind of physical sense? Certainly if you add more and more terms of the above series, you are not getting in any sense close to $-1$. And at the crude informal level, it is clear that the sum is "infinite," whatever that may mean. Formal manipulations on non-existent objects can yield absurd results.

So in this case formal manipulation has produced a result that makes no physical sense. There are a number of instances where blind manipulation yields wrong results, so early in university, students are trained to apply algebraic "rules" only in situations where these rules are valid.

However, the situation is more complicated than that! Once upon a time complex numbers, when they came up in a formal calculation, were dismissed as "absurd." Now they are part of the essential toolkit of the electrical engineer! There are similar phenomena with series.

In combinatorics, for example, formal power series are successfully used, at the purely manipulational level, even when the series technically do not converge. And in various areas of mathematics, divergent series of a suitable type can be very useful, even in computations!

But at this stage, you should be disciplined, and apply formal manipulations only in situations where you know, or have been assured that, they are "safe." Later, perhaps, you can explore situations where venturing beyond safe confines yields interesting and useful results.

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I gave a +1 especially because of the last paragraph. – J. M. May 14 '11 at 6:46
Thanks a lot, but i have a question of formal manipulations making absurd results. What was the thing that leads this to non-sense? Is it subtracting 1 from both sides which leads to S-1=rS? or is it the isolating the S with an assumption of existence of it's value which is wrong? – Jee Hoon Park May 14 '11 at 7:17
The answer might be that S-1=rS does have a solution when r is 1 or greater, only this solution is S=+infinity and that in the real line extended by +infinity there is no substraction, only addition. In other words the step where everything breaks down is not S-1=rS nor S=rS+1 (both perfectly legal) but S-rS=1. – Did May 14 '11 at 7:46
@Jee Hoon Park: In a way, the first (deliberate) mistake was to call the sum $S$, as if it actually existed. But you can also think of the error in the calculation by saying the sum "really is" $\infty$, and we are saying $\infty - 1 = (2)(\infty)$, obviously wrong since we cannot freely use the algebra of ordinary numbers on "infinity" (whatever that may mean). However, as I pointed out, divergent series really can be useful, if you know what you are doing. – André Nicolas May 14 '11 at 7:52
thanks, I couldn't find any clear explanation of this absurd result in my highschool text book haha now I'm ok with it – Jee Hoon Park May 14 '11 at 7:55

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