Today, a friend gave me a "proof" of $1=2$ and challenged me to find the fallacy.

$1 = 1$

$1 = 1 + 0 + 0 + 0 ...$

$1 = 1 + 1 - 1 + 1 - 1 + 1 - 1 ...$

$1 = 2 - 1 + 1 - 1 + 1 - 1 ...$

$1 = 2 + 0 + 0 ...$

$1 = 2$

My answer was that once you turn the initial $1 + 1$ into a 2, everything is offset so a $-1$ is always left at the end no matter how many times it is repeated. This negative one balances out the $2$ at the beginning so $1=1$ still holds true. I.e.

$$1 = 1 + 0 + 0 + 0 ... = 1 + (1 - 1) + (1 - 1) + (1 - 1) = 2 + (-1 + 1) + (-1 + 1) - 1$$

However, my friend claimed that my answer only applies if the $+ 1 - 1$ repeats for a finite number of times. He argues that because the sequence repeats infinitely and things work differently when working with infinity, my answer is not valid.

Can anyone enlighten me to the true fallacy in this proof?

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    $\begingroup$ I don't have the time to give an answer right now and I'm sure someone will, but there's one thing you should think about first: what does the symbol $1+1-1+1-1+1-1\cdots$ really mean? There's a rigorous way to define it and that's exactly where the problem lies. Infinity takes a huge role on this matter. So that people can better help you, are you familiar with the concept of series? $\endgroup$ – Git Gud Jun 20 '13 at 7:18
  • $\begingroup$ I do have a basic understanding of series. $\endgroup$ – tangrs Jun 20 '13 at 7:24
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    $\begingroup$ I had similar question about that: math.stackexchange.com/questions/417280/… $\endgroup$ – Mher Jun 20 '13 at 7:24
  • $\begingroup$ Thanks for that, I think my question may actually be a duplicate of that. $\endgroup$ – tangrs Jun 20 '13 at 7:26
  • $\begingroup$ I think the truth reason is "addition of an infinite number of terms cannot be assumed to be associative or commutative. In particular, rearranging terms in an infinite sum can actually change the sum." says here math.stackexchange.com/questions/427635/… $\endgroup$ – iMath Apr 19 '15 at 8:39

Some users pointed me to read up on converging and diverging series.

As I currently understand it, equating $1+0+0+0... = 1+1−1+1−1+1−1...$ is the fallacy because the series on the right does not converge (much less to 1) - therefore, they are not equal.

To prove that this is the fallacy, we can use convergent tests to show that the two sides of the equation are not equal.

Am I correct in my deduction?

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    $\begingroup$ Pretty much so. The symbol $1+1−1+1−1+1−1\ldots$ is just a different way of writing $\sum \limits_{n=0}^{+\infty}a_n$, where $a_0=1$ and $a_{n+1}=(-1)^n$, for all $n\in \Bbb N_0$. And now what does $\sum \limits_{n=0}^{+\infty}a_n$ mean? Well, that's just $$\sum \limits_{n=0}^{+\infty}a_n=\lim \limits_{m\to+\infty}\left(\sum \limits_{n=0}^{m}a_n\right)$$ if the limit exists, but this limit doesn't exist, so it doesn't make sense to talk about the series. A simple way to see the series doesn't converge is to note that $(a_n)_{n\in \Bbb N_0}$ doesn't converge to $1$, (it does not converge). $\endgroup$ – Git Gud Jun 20 '13 at 7:58
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    $\begingroup$ Thanks, that explains it perfectly $\endgroup$ – tangrs Jun 20 '13 at 8:09
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    $\begingroup$ I meant doesn't converge to $0$, above. $\endgroup$ – Git Gud Jun 20 '13 at 14:30

We need to discuss two related things here: what mathematicians (as opposed to the general public) mean by "$\dots$", and how we produce a way of working with infinite sums that both makes sense, and is consistent.

What does "$\dots$" mean?

When the layperson writes "$\dots$", they normally mean either

  • "and so on, until the obvious logical endpoint", or
  • "and so on, forever"

But mathematicians need something more precise:

  • The first usage is written mathematically by writing the endpoint afterwards: $a_1+a_2+\dotsb+a_n$ simply means to add all the numbers in the sequence with labels between $1$ and $n$. Provided we know what all these are, this is straightforward: for example, $ 1+2+\dotsb+100 = 5050 $.
  • The second usage is a problem: $a_1+a_2+\dotsb$ has no last member, so we need to add infinitely many things.

How do we add infinitely many things?

The answer is that we don't. This sounds silly, but first ask yourself this: how do you add three things together? It's impossible to add three numbers at once, so what you do is add two, and then add the third to the result. Similarly for $n$ things.

What about infinitely many? The clue lies in the previous paragraph: we compute $$\begin{align} s_1 &= a_1 \\ s_2 &= a_1 + a_2 = s_1 + a_2 \\ s_3 &= a_1 + a_2 + a_3, \end{align} $$ and so on, using the rule that $s_n = s_{n-1} + a_n$. $s_n$ are called the partial sums.

But this is still a load of finite sums. How do we assign a value to the whole infinite sum?

The Limit of the Partial Sums

We say that $a_1+a_2+\dotsb$ has a sum (or is summable) if $s_n$ tends to a limit, say $s$, as $n$ tends to infinity. In particular, we can make $s_n$ as close to $s$ as we like by taking $n$ large enough. In particular, it is therefore required that the terms in the series converge to zero, since $s_{n}-s_{n-1}=a_n$. This is obviously not the case for Grandi's series.

The major advantages of this definition include that it assigns a unique sum to $a_1+a_2+\dotsb$,

  1. $ ka_1 + ka_2 + \dotsb = ks $, and
  2. if $a_n'=a_{n+1}$, $a'_1 + a'_2 + a'_3 + \dotsb = s-a_1 $.

How, therefore, does one manipulate these objects that look like series, but do not have a sum?

Assigning values to divergent series

We would like a method to give divergent series a consistent meaning so that we can manipulate them in a meaningful way: i.e., so we find unique values, compatible with previous theory, and that we can apply simple algebraic operations to. We write $S(a_n)$ to be the value assigned to $a_1+a_2+\dotsb$ We normally therefore ask that a summation method has

  1. Scaling: $ S(ka_n) = kS(a_n) $,
  2. Addition: $ S(a_n+b_n) = S(a_n)+S(b_n) $,
  3. Relabeling: $a_1+S(a_{n+1}) = S(a_n)$ (not the standard name, but never mind),
  4. Regularity: If $a_1+a_2+\dotsb$ is convergent, $S(a_n)$ agrees with the conventional sum,

and I think most people would agree that all of these are sensible things you want a summation method to do.

Supposing that we demand that Properties 1. and 3. hold for our summation technique. Then Grandi's series satisfies $$ s = 1-1+1-1+\dotsb = 1-(1-1+1-1+\dotsb) = 1-s, $$ so any "reasonable" method of summation gives Grandi's series the value $s=1/2$.

For much more information on this, I recommend G.H. Hardy's definitiv work on the subject, Divergent Series. In particular, Chapter 1 discusses this and other series, and the four properties given above are in § 1.3.


The third equation your friend presented to you is nonsense. The sum on the right does not converge.


I'm sorry for posting this late, I just wrote this up as an answer for this (Why is this $0 = 1$ proof wrong?) question before it was marked duplicate. Maybe it has some tiny new bits in it

One possible answer: The proof is wrong because infinite summation is not associative (as is for example shown by this :D )

It is however interesting because it shows that when dealing with infinite series, one has to be careful what one is actually talking about and therefore introduce some notions that might seem technical at a first glance. For example there is usually made a distinction between the sequence of partial sums and it limes.

If the latter one exists (i.e. if the series converge), one sometimes identifies it with it's series, but this identification can lead to problems as in the argument you show: In the first and second line, you are dealing with convergent series, so you can make the identification of limes (left hand side) and sequence of partial sums (right hand side). In the third line, the only way to make sense of the right hand side is a sequence of partial sums, the left hand side is a number however, so we cannot write down equality. The last lines have again a convergent sum on the right, but the sequence of partial sums is different from the one in line two!

I hope this was not too technical, feel free to ask for explanations!

By the way, it is very interesting, that in other contexts, precisely the above argument can be used to show interesting (true) things, for example that you cannot unknot a knot by adding another one to it (check http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle if you are interested)

Also the proofs of the "sum of natural numbers equals minus one over 12" proofs you mention are probably wrong (Is it the youtube videos?) and i would say not accepted fact (and i really mean the proofs, the statement can still be made sense of in a different way) but still a good way to generate interest for these questions i guess :)


Let's acknowledge that the sequence $1,0,1,0,1,0,...$ is divergent. The result will follow from this fact. Now, consider the series $$\sum_{n=1}^{\infty}(-1)^{n-1}=1-1+1-1+\ldots\:\:\:\:\:\:\:\:\:\:\:\:(*)$$ which appears on your calculation. By definition, we know that the limit of the sequence of partial sums is the limit of the series. Sequence of the partial sums of our series is as follows:

$$(s_n)=1,0,1,0,1,0,\ldots$$ which is divergent as we know and indicated in advance.

Therefore, treating the RHS of the equation $(*)$ as a number is mathematically incorrect, so calculations may yield wrong results.


Here's my answer!

This example gives you the answer $\displaystyle \frac{1}{2}$ , but it doesn't follow the rules of this taylor series formula

$\displaystyle \begin{array}{{>{\displaystyle}l}} \sum ^{\infty }_{n=0} x^{n} =\frac{1}{1-x} ,\ | x| < 1\Longrightarrow \\ \sum ^{\infty }_{n=0}( -1)^{n} =\frac{1}{2} \end{array}$

This one was done by Srinivasa Ramanujan

Ramanujan summation:

$\displaystyle \begin{array}{{>{\displaystyle}l}} \sum\limits ^{\infty }_{n=1} f( n)\overset{\Re }{=} -\frac{f( 0)}{2} +i\int\limits ^{\infty }_{0}\frac{f( ix) -f( -ix)}{e^{2\pi x} -1} dx\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathfrak{Grandi's\ series}\\ \sum ^{\infty }_{n=0}( -1)^{n} =1+\sum\limits ^{\infty }_{n=1}( -1)^{n}\overset{\Re }{=}\\ 1-\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{-ix}}{e^{2\pi x} -1} dx=\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{-ix}}{e^{2\pi x} -1} dx\\ My\ friend\ Nico\ figured\ this\ one\ out:\ ( -1)^{a} =( -1)^{-a} \Longrightarrow \\ \frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{ix}}{e^{2\pi x} -1} dx=\frac{1}{2} +i\int\limits ^{\infty }_{0} 0dx=\frac{1}{2} \Re \end{array}$

\begin{array}{l} Note\ that\ the\ Ramanujan\ summation\ does\ a\ partial\ sum,\ and\ assignes\ a\ value\ to\\ an\ otherwise\ divergent\ series.\ This\ does\ not\ mean\ that\ the\ actual\ answer\ to\\ Grandi's\ series\ is\ \displaystyle \frac{1}{2} .\\ \\ Abel's\ theorem\ and\ regularization\ say's\ it\ works\ also,\ as\ well\ as\ borel\ regularizations.\\ \\ Ernesto\ Cesàro's\ try\ at\ it \end{array}

\begin{array}{l} For\ a\ divergent\ sum\ \displaystyle S=\sum ^{\infty }_{n=1} a_{n} ,\ let\ s_{k} \ be\ the\ partial\ sum:\ s_{k} =\sum ^{k}_{n=1} a_{n}\\ The\ sequence\ is\ called\ Cesàro\ summable\ if\ as\ \displaystyle k \ tends\ to\ infinity\\ the\ arithmetic\ mean\ of\ it's\ first\ \displaystyle k \ partial\ sums\ tends\ to\ the\ value\ \displaystyle A\in \mathbb{R} :\\ \displaystyle \lim _{m\rightarrow \infty }\left(\frac{1}{m}\sum ^{m}_{k=1} s_{k}\right) ,\\ let\ \displaystyle G \ denote\ Grandi's\ series,\ let\ \displaystyle ( s_{k})^{\infty }_{k=0} \ denote\ the\ sequence\ of\ partial\ sums\ of\ \displaystyle G .\\ even\ though\ this\ diverges\ \displaystyle G \ is\ Cesàro\ summable.\\ Let\ \displaystyle ( t_{m})^{\infty }_{m=1} =\frac{1}{m}\sum ^{m}_{k=1} s_{k}\\ \displaystyle t_{m} =\left(\frac{1}{1} ,\frac{1}{2} ,\frac{2}{3} ,\frac{2}{4} ,\frac{3}{5} ,\frac{3}{6} ...\right) \Longrightarrow \lim _{m\rightarrow \infty }( t_{m}) =\frac{1}{2}\\ As\ \displaystyle m \ approaches\ infinity,\ the\ terms\ of\ \displaystyle t_{m} \ tend\ toward\ \displaystyle \frac{1}{2} .\\ \\ Because\ many\ methods\ of\ looking\ at\ this\ sum\ give\ you\ the\ value\ \displaystyle \frac{1}{2} ,\ and\ because\\ it\ seems\ like\ a\ good\ answer\ (because\ it\ oscillates\ between\ 0\ and\ 1)\ people\\ say\ that\ \displaystyle \frac{1}{2} \ is\ the\ best\ answer \end{array}

Still of course, Grandi's series is considered divergent. I actually have a page on my website called trig(infinity) that goes a little more in depth on Grandi's series- Here's the link!

Basically you can say that the value $\frac{1}{2}$ is Grandi's series assigned value I hope this helps!


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