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We know that: https://www.youtube.com/watch?v=w-I6XTVZXww $$S=1+2+3+4+\cdots = -\frac{1}{12}$$

So multiplying each terms in the left hand side by $2$ gives: $$2S =2+4+6+8+\cdots = -\frac{1}{6}$$ This is the sum of the even numbers

Furthermore, we can add it to itself but shifting the terms one place: $$ \begin{align} 1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\ =1+3+5+7+\cdots & =2S \end{align} $$ This is the sum of the odd numbers

If we were to now sum the odd numbers and the even numbers like below: $$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$ This supposedly tells us that: $$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$

What is faulty with this proof.

Michael Hardy
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user3184807
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    The first line is false. – Travis Willse Feb 13 '16 at 16:50
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    google it https://www.youtube.com/watch?v=w-I6XTVZXww – user3184807 Feb 13 '16 at 16:50
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    You can't manipulate series this way (unless they converge absolutely in the traditional sense). You don't need such an abstract sum to get apparent paradoxes...if you let $T=1-1+1-1+\dots$ then it is easy to write arguments similar to yours to show that $T=0,1,\frac 12, \dots $. – lulu Feb 13 '16 at 16:52
  • why not, surely shifting the numbers is fine – user3184807 Feb 13 '16 at 16:52
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    Google it : https://en.wikipedia.org/wiki/Ramanujan_summation (Although the Ramanujan summation of a divergent series is not a sum in the traditional sense... ) – Yiyuan Lee Feb 13 '16 at 16:53
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    No, it isn't. With $T$ as I defined it we can easily get $T=(1-1)+(1-1)+\dots=0$ or $T=1-(1-1)-(1+1)\dots=1$. – lulu Feb 13 '16 at 16:54
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    I'm familiar with the statement, it's not literally true as written. What is true, for example, is that there's a natural way to extend the function $\zeta(s) := \sum_{k = 1}^{\infty} k^{-s}$, which is defined on $(1, \infty)$ to most complex numbers, including $s = -1$, and $\zeta(-1) = -\tfrac{1}{12}$. The partial sums of the series, evaluated at $s = -1$, are $1 + \cdots + n$, but this is not the same as saying $1 + 2 + 3 + \cdots = -\tfrac{1}{12}$. – Travis Willse Feb 13 '16 at 16:54
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    Even conditionally convergent series can be shifted around to get different results. See Riemann series theorem. – JMoravitz Feb 13 '16 at 16:54
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    In fact, for any real number $a$, one can rearrange the terms in any conditionally but not absolutely convergent real series so that the resulting series converges to $a$. – Travis Willse Feb 13 '16 at 16:56
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    If the first line can be said to be true, it is only in the context of a certain "summation method", and all the subsequent reasoning would have to be consistent with that. $\qquad$ – Michael Hardy Feb 13 '16 at 16:59
  • A related question. – Lucian Feb 13 '16 at 21:53

3 Answers3

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Interpreted literally (i.e., using the usual sense of limits of infinite series), the first line, $$1 + 2 + 3 + \cdots = -\frac{1}{12} ,$$ is simply false, as the series on the l.h.s. diverges.

What's true, for example, is that there's a natural way to extend the function $$Z(s) := \sum_{k = 1}^{\infty} k^{-s} ,$$ which is defined on $(1, \infty)$ (and in particular not at $s = -1$), to a function $\zeta$ defined for most complex numbers, including $s = -1$, and this function satisfies $\zeta(-1) = -\tfrac{1}{12}$. The partial sums of the series, evaluated at $s = -1$, are $1 + \cdots + n$, but this is not the same as saying $1 + 2 + 3 + \cdots = -\tfrac{1}{12}$.

Travis Willse
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As pointed out in section 3, page 1191 of this article, the rules for manipulating divergent series are more restrictive than those for convergent series. As pointed out in the article, to avoid problems you should work with the power series obtained by multiplying the nth term by $x^n$, instead.

Count Iblis
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Lets try the same in a more general way: Instead of $-\frac{1}{12}$ I will use $C\in\mathbb{R}$.

So our infinite sum: $$1+2+3+4+\cdots = C$$

Using the same technique you get these two equalities:

$$ 2+4+6+8+\cdots = 2C \\ 1+3+5+7+\cdots = 2C $$

And when you add them together:

\begin{align} 1+2+3+4+\cdots &= 4C \\ C &= 4C \end{align}

So for every nonzero real number $C$ you get $1 = 4$ which is obviously not true. So to answer your question: Why is the proof faulty for C = $-\frac{1}{12}$?? Because it's faulty for every nonzero real number. The mistake is, as it's been already said, in the first line. You can't prove something that is wrong. Learn more about this infinite series here.

Jeyekomon
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