A common "trick" for obtaining a closed form of a geometric series is to define $$ R := \sum_{k=0}^{\infty} r^k, $$ then manipulate the series as follows: \begin{align} R - rR &= \sum_{k=0}^{\infty} r^{k} + \sum_{k=0}^{\infty} r^{k+1} \\ &= (1 + r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\ &= 1 + (r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\ &= 1. \end{align} On the other hand, $R-rR = (1-r)R$. Hence $$ (1-r)R = 1 \implies R = \frac{1}{1-r}. $$ In this example, the formula is obtained by a sequence of relatively elementary algebraic manipulations.

By a similar kind of manipulation, suppose that $$ S := 1 + 1 + 1 + 1 + \dotsb = \sum_{k=0}^{\infty} 1. $$ $S$ is unaffected by addition of $1$, and so $S = 1+S$. Canceling $S$ from both sides gives $0 = 1$, which is clearly nonsense.

Question: What went wrong with the second computation? Why do these arguments work well for summing the geometric series, but not for the series of ones?

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    $\begingroup$ I am posting this question specifically as a dupe / merge target for a couple of low quality questions which have been the source of some controversy recently: (1) and (2) $\endgroup$ – Xander Henderson Sep 12 '19 at 17:53

To understand things like this, you have to pay careful attention to the underlying definitions. The definition of an infinite sum, like

$$1 + 1 + 1 + 1 + \cdots$$

is the limit

$$\lim_{n \rightarrow \infty} \underbrace{1 + 1 + \cdots + 1}_{n}$$

i.e. the sum of $n$ ones, as $n$ is allowed to approach infinity. However, this limit does not exist in the real number system, because the right-hand term grows indefinitely large.

Yet, by substitution, this limit is the value you have decided to represent by the symbol $S$. Your problem, then, is that such a value does not exist. The sum of the infinite series doesn't exist. Hence $S$ has no referent, and the associated computations are meaningless.

That said, an alternative, and perhaps stronger, perspective would be to say that if an object like $S$ existed, and it permitted the manipulations you did, it would break things, because its existence would thus embody contradictions.

Of course you may be wondering, then, "but what about $\infty$? Isn't

$$\lim_{n \rightarrow \infty} \underbrace{1 + 1 + \cdots + 1}_n = \infty$$


The answer is: no, not in the real number system. In the real number system, the limit does not exist. The above equation is often shown, but its meaning is not really made clear. What it "really" means is an equation in the extended real number system, where an additional element called $\infty$ has been added, and that results in the prior limit as being valid. In that case, then yes, $S = \infty$. Yet, given the last paragraph of what I just said above, something has to break for this not to be contradictory. What breaks is that $\infty$, as an extended real number, but not a real number. And once allow $S$ to take extended-real values, the very rules of algebra change, as you are working in a different number system - it is like going into the complex numbers by adding $i$. Namely, in the extended real numbers you are not allowed to start with

$$S = 1 + S$$

then "subtract from both sides"

$$S - S = (1 + S) - S$$

and then "cancel". The subtraction is okay, but not the cancellation. You now cannot infer that the left-hand side is zero. In fact, $\infty - \infty$ is, itself, undefined, in this extended real number system.

If you go this route, what you learned in grade school quits working.

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    $\begingroup$ Or another interpretation is that $\lim_{n->\infty} f(n) = \infty$ is a special notation which means something entirely different from two things being equal. $\endgroup$ – aschepler Aug 17 '19 at 1:06
  • $\begingroup$ @aschepler : Yes, but I prefer things that are more unified, consistent, and cogent. Instead of three definitions of limit on one space, you can have two spaces and one cogent definition of limit in terms of open sets. Moreover, it is very often useful to treat $\infty$ as a pluggable object, so by getting the logic and the rules for that mentioned and set for that up front, you avoid confusing crap like "this thing isn't a number but we're gonna use it like one anyways". I don't like that. $\infty$ is a different kind of number, and a very, very useful one (esp if, say, you're working with $\endgroup$ – The_Sympathizer Aug 17 '19 at 1:36
  • $\begingroup$ logarithmic scales like decibels where you need it to represent a value of zero) - you just have to know them ropes and treat him the right way so he don't bite you :) Good lil puppdog, good lil puppiedoggie :) #mehhr. $\endgroup$ – The_Sympathizer Aug 17 '19 at 1:36
  • $\begingroup$ You can also think of it like this. If you have the equation $x+1 = x$, you get $1=0$. Did you do anything amazing? No, because the starting equation was obviously false. Also, if you have $S = S+1$ where $S$ is the sum of a diverging series, that is basically like subtracting $\infty$ from both sides, and $\infty - \infty = 0$ is a false statement. $\endgroup$ – Some Guy Feb 25 at 22:53

Your argument hinges upon the assumption that $S$ is a number you can do arithmetic with. It isn't a number, you can't do arithmetic with it, and this is what you have shown (by contradiction).

  • $\begingroup$ @samerivertwice I am pretty certain most people, when reading this question, accept that we are talking about the conventional real (or perhaps complex) numbers, and nothing else. Taking into account all manner of pathological algebraic structures when writing a response to this question is just a pedantic hassle. I think that this answer is not problematic in that respect whatsoever. $\endgroup$ – Arthur Mar 15 at 21:07
  • $\begingroup$ I don't consider $\Bbb Q/\Bbb Z$ or $1+\omega$ as especially pathological when the domain has not been declared. $\endgroup$ – samerivertwice Mar 22 at 11:46
  • $\begingroup$ @samerivertwice When it is tagged real analysis, and not a single word of the question hints otherwise, yes, those are pathological. $\endgroup$ – Arthur Mar 22 at 12:06
  • $\begingroup$ 👍🏼 I can see I'm not going to win you over. I was aware of the real analysis tag but no proof that $1=0$ is possible in real analysis. The question's identified as a target for duplicate similar questions, whereas many of those pointed here probably won't be tagged real analysis. $\endgroup$ – samerivertwice Mar 22 at 13:14

You are treating infinity as if it were a number. However, it is not, and therefore you cannot perform ''usual'' operations such as $+$ and $\times$ on it.

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    $\begingroup$ The problem (at least among positive numbers) is not $+$ and $\times$ but $-$ and $\div$ $\endgroup$ – Henry Aug 16 '19 at 12:15

Since the expression $1+1+\cdots$ makes no sense as a number, there is nothing that you can prove from it using algebraic computations.



The basic problem is that the object $S$ defined in the question is nonsense, at least within the scope of "normal" mathematical discourse. Thus the question really comes down to "Why is $R$ well-defined while $S$ is not?" The answer to this question comes down to definitions.

Definition: Given a series of the form $$ \sum_{k=1}^{\infty} a_k, $$ where each $a_k$ is a real number, define the $n$-th partial sum by $$ T_n := \sum_{k=0}^{n} a_k. $$ We say that the original series converges to a real number $T$ if the partial sums converge to $T$ as $n$ goes to infinity. That is, the series converges to $T$ if $$ \lim_{n\to \infty} T_n = T. $$ In this case, we write $$ T = \sum_{k=0}^{\infty} a_k. $$ If a series does not converge to a finite limit, then we say that it diverges.

The Geometric Series

When working with a geometric series, we can obtain the result in the question directly from the definition. In that case, the partial sums are given by $$ R_n = \sum_{k=0}^{n} r^k. $$ The goal is to compute $\lim_{n\to\infty} R_n$, which can be done by first finding a useful closed form for each $R_n$. This can be done by paralleling the computations in the question, but in a way that can be justified rigorously. For each fixed $n$, we have \begin{align} (1-r)R_n &= R_n - rR_n \\ &= \sum_{k=0}^{n} r^k - \sum_{k=0}^{n} r^{k+1} \\ &= (1 + r + r^2 + \dotsb + r^{n-1} + r^n) - (r + r^2 + r^3 + \dotsb + r^n + r^{n+1}) \tag{1} \\ &= 1 + (r + r^2 + \dotsb + r^n) - (r + r^2 + \dotsb + r^n) - r^{n+1} \tag{2} \\ &= 1 + r^{n+1}. \end{align} At (1), we are just expanding out the notation. At (2), we are using the fact that addition is associative, and so we can move the parentheses around at will.[1] This computation then gives $$ (1-r)R_n = 1 - r^{n+1} \implies R_n = \frac{1-r^{n+1}}{1-r}. $$ As long as $r \ne 1$, this formula for the $n$-th partial sum is perfectly well-defined. If $r = -1$, then this expression oscillates between $\frac{1}{2}$ and $-\frac{1}{2}$, depending on the parity of $n$. Finally, if $|r| > 1$, then the magnitude of the numerator grows without bound, and the sequence of partial sums fails to converge. Otherwise, i.e. if $|r| < 1$, we can take a limit to get $$ \lim_{n\to\infty} R_n = \lim_{n\to\infty} \frac{1-r^{n+1}}{1-r} = \frac{1}{1-r}. $$ Therefore, from the definition of a convergent series, we are justified in writing $$ \sum_{k=0}^{\infty} r^k = \frac{1}{1-r}, $$ assuming that $|r| < 1$. The "algebraic manipulations" in the question are, in a sense, a shortcut through this more formal computation.

The Series of Ones

In the case of the series of ones, things go wrong. The $n$-th partial sum is given by $$ S_n = \sum_{k=0}^{n} 1 = n+1. $$ But then $$ \lim_{n\to\infty} S_n = \lim_{n\to\infty} (n+1) = \infty. $$ The sequence of partial sums is unbounded, and therefore does not converge to a real number. In other words, the series $$ \sum_{k=0}^{\infty} 1 = 1+1+1+1+\dotsb $$ cannot reasonably be assigned a real value. Since it cannot be assigned a real value, further algebraic manipulation is meaningless. Further discussion of this problem can be found in the answers to this question about arithmetic with infinite quantities.

[1] Since we are working with infinite series here, it is important to note that finite addition is associative. That is, if we have a finite number of terms which we want to add together, we can rearrange the parentheses however we like. This does not work with an infinite number of terms. See, for example, Grandi's series.


I found infinity a lot easier to deal with once I understood what I consider a very simple truth.

"Normal numeric operations simply don't work right if you try to apply them to infinity."

You've already found one contradiction that illustrates this. Here's another.

$\infty + \infty = \infty$

Subtract $\infty$ from both sides and we have:

$\infty = 0$


In fact, for $|r|<1$,

$$R=\lim_{n\to\infty}\sum_{k=0}^n r^k$$ and

$$(1-r)R=\lim_{n\to\infty}\left(\sum_{k=0}^n r^k-\sum_{k=0}^n r^{k+1}\right)=\lim_{n\to\infty}(1-r^{n+1})=1.$$ All steps are rigorous.

If we try a similar argument with


we obtain

$$S+1=\lim_{n\to\infty}\sum_{k=0}^{n+1}1$$ and by substraction,

$$S+1-S=\lim_{n\to\infty}\left(\sum_{k=0}^{n+1}1-\sum_{k=0}^n1\right)=\lim_{n\to\infty} 1\ne0.$$

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    $\begingroup$ You get to the right result in the end but, given that neither of the first two limits exists, your passing of the limit through the sum is not justified. $\endgroup$ – Xander Henderson Mar 15 at 21:21
  • $\begingroup$ @XanderHenderson: this is precisely what I am showing: if the computation is done without justification, the results are wrong. Apparently, my readers understood differently. $\endgroup$ – Yves Daoust Apr 2 at 19:29

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