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At the end of the Wikipedia article on:

$$1+2+3+4 +\dots$$

an argument is present that the sum adds up to $-\frac{1}{12}$.

Here is the numberphile video.

Here's my attempt to fill in the details from the argument:

(1) Let $S_1 = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots$

(2) Let $S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$

(3) $2S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\, 1 - 2 + 3 - 4 + 5 - 6 + 7 + \dots$

(4) $\,\,\,\,\,\,\,\,\,\, = 1 -1 + 1 -1 + 1 - 1 + 1 - 1 + \dots$

(5) $S_1$ is the Grandi's series whose Cesaro's sum is $1/2$

(6) So, $S_2 = 1/4$

(7) Let $S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$

(8) $S - S_2 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 + 2 - 3 + 4 - 5 + 6 -7 + 8 + \dots$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 0 + 4 + 0 + 8 + 0 + 12 + 0 + 16 + \dots$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 4 + 8 + 12 + 16 + 20 + 24+ 28 + \dots$

(9) $4S = 4 + 8 + 12 + 16 + 20 + 24 + 28 + \dots$

(10) $S - S_2 = S - 1/4 = 4S$

(11) Solving for $S - 1/4 = 4S$ gives $S = -\frac{1}{12}$

Is this argument valid? Is there a flaw in the logic that leads to the right answer in the wrong way?

Larry Freeman
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    Not valid. The sum is $\infty$. That is, $\sum_{i=1}^{\infty} 1 = \infty$. – Michael Nov 05 '15 at 23:52
  • I am especially interested in validity of the argument by the NumberPhile video which I do not believe is mentioned in the other question. – Larry Freeman Nov 05 '15 at 23:52
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    The $S_1$ value you give is not well defined. Its sum oscillates between 0 and 1 and does not converge to anything. It does not make sense to treat it as if it were a real number. – Michael Nov 05 '15 at 23:53
  • Sums of the form $\sum_{i=1}^{\infty} a_i$ are defined as limits of the "partial sums" $\sum_{i=1}^n a_i$ as $n\rightarrow\infty$, and are defined only when the limit makes sense. That is, by definition, $\sum_{i=1}^{\infty} a_i = \lim_{n\rightarrow\infty} \sum_{i=1}^n a_i$, when the limit makes sense. – Michael Nov 05 '15 at 23:54
  • @Michael Not always... – Michael Biro Nov 05 '15 at 23:55
  • @MichaelBiro : Care to clarify? – Michael Nov 05 '15 at 23:55
  • @Michael Different definitions are possible and interesting to explore. See Abel, Cesaro, or Ramanujan summation... – Michael Biro Nov 06 '15 at 00:00
  • Well if you want to change the definition then it would help to fix that at the beginning, especially if it conflicts with the standard definition. It is misleading otherwise. @MichaelBiro – Michael Nov 06 '15 at 00:02
  • Sums of nonnegative numbers are easiest to deal with. – Michael Nov 06 '15 at 00:03
  • @MichaelBiro : I once wasted time watching a $\sum_{i=1}^{\infty} 1 = -1/12$ video, perhaps the same one by the asker. I don't think the people in the video even knew there were such things as "definitions." – Michael Nov 06 '15 at 00:06
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    @Michael I agree wholeheartedly. The Numberphile video is unfortunately extremely misleading, trying to shock people with an intuitively ridiculous result that has (other) abstract justifications. – Michael Biro Nov 06 '15 at 00:07
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    Voting to reopen, as the "duplicate" does not actually address the logical fallacy of the alleged proof, instead focusing on another argument for the same result. – mweiss Nov 06 '15 at 00:09
  • @Eric, thanks for the link. I didn't see the other question. Should I delete this one? How should I handle a duplicate? – Larry Freeman Nov 06 '15 at 00:25
  • @LarryFreeman : If you want to define $S_1$ in terms of a Cesaro average $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n (-1)^{i+1}$ then it is 1/2, but you will have problems defining $S_2$ in terms of a Cesaro average since even its Cesaro average does not have a well defined limit! Even if it did, you might have problems mixing Cesaro averages with regular definitions of sums. Overall you need to be precise up front about what your definition is. – Michael Nov 06 '15 at 00:28
  • Thanks, @MIchael. I was pretty sure that the proof was invalid. Thanks for your explanation. I wanted to be clearer on why it was invalid. Your explanation helps. – Larry Freeman Nov 06 '15 at 00:30
  • @LarryFreeman: Now that it has been reopened, I can mark it as a duplicate of the other question and you should get an option to agree that it is a duplicate. – Eric Wofsey Nov 06 '15 at 00:41
  • Aww. I remember that video. It is what first got me interested in number theory. –  Nov 06 '15 at 00:54
  • I've just posted an answer to a similar question which covers also the motivation and minimal requirements so that we don't make a simple error. I think it could be helpful for your question too: http://math.stackexchange.com/a/1604151/1714 – Gottfried Helms Jan 08 '16 at 11:32

2 Answers2

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The argument is invalid for one simple reason:

The operator "$+$" is not defined for infinite sums.

Let me elaborate:

The symbol $+$ is a symbol that means something to us. It is a symbol that means "take whatever whole/rational/real/complex number is to the left of me and add it to whatever number is to the right of me."

In mathematical terms, $+$ is a mapping from $X\times X$ to $X$, i.e., if $X=\mathbb R$, then $+$ is a mapping that maps the pair $(x,y)$ into $x+y$. This would be more clear if we would write $$+(5,10) = 15,$$ but for historic reasons, we write $$5+10=15$$

Now, $+$ has some properties. For example, we know that $a+(b+c) = (a+b)+c$ for any three numbers $a,b,c$. This means that for any finite series of numbers $a_1,a_2,\dots a_n$, we know that no matter what the order in which we add the numbers, their sum is the same. That's why we are allowed to simply write

$$a_1+a_2+\dots + a_n$$

to denote the sum of the numbers.

However there is no trivial way in which we can define the sum of an infinte number of numbers!!!

The standard definition of a sum of real numbers is:

$$a_1 + a_2 + a_3 + \cdots = \lim_{n\to \infty} (a_1 + a_2 + \cdots + a_n),$$

and using this definition, the sum of all ones is $\infty.$

5xum
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We need to be very careful with blind manipulations like the ones in the video - in particular, you can get situations like:

$$S = 1 + 2 + 3 + 4 + \dots$$

$$-S = S - 2S = (1 + 2 + 3 + 4 + \dots) - (2 + 4 + 6 + 8 + \dots)$$ $$= 1 + (2 - 2) + (3 - 4) + (4 - 6) + (5 - 8) + \dots = 1 + 0 - 1 - 2 - 3 - 4 +\dots = 1 - S$$

So $-S = 1- S$ and $0 = 1$.

Michael Biro
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