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Here is how I derived this...

$$1+2+3+4+...=-\frac{1}{12}$$ $$2+4+6+8+...=2(1+2+3+4+...)=2(-\frac{1}{12})=-\frac{1}{6}$$ Thus $1+3+5+7+...=-\frac{1}{6}-(1+1+1+...) $ because $S_{odd}=S_{even}-(1+1+1+...)$

Also $S_{odd}=(1+2+3+4+...)-(2+4+6+8+...)=-\frac{1}{12}+\frac{1}{6}=\frac{1}{12}$

Thus $\frac{1}{12}=-\frac{1}{6}-(1+1+1+...)$ $$1+1+1+...=-\frac{3}{12}$$

What I think my mistake was (if I have one) is where I assume the sum of all numbers is half the sum of all even numbers; although, it should work since there is nothing two times infinity.

Please leave simple solution (im only 15 years old) to why this is false: Wikipedia says that $\sum_{n=1}^{\infty} 1 = \infty$.

Edit: I know now that the above notion is false. I recently watched a Numberphile video proving $\sum_{n=1}^{\infty} n = -\frac{1}{12}$. I followed the same line of reasoning to derive the untrue $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ I'll admit; I was ignorant in believing so and I apologize for wasting your time.

Thanks for the answer kindly explaining why I was wrong. I guess I should watch this video to un-brainwash me. Sorry again.

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    $\begingroup$ While I can understand why one would be tempted to downvote this question (with so many fundamental conceptual errors), from the perspective of the OP this shows research effort and I would argue might represent what a large number of amateurs might think after watching, say, the infamous numberphile video on $\sum n=-1/12$. For what it's worth, I think this is a good question, at least in this sense. $\endgroup$ – YiFan Oct 5 at 22:33
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    $\begingroup$ PS There's nothing to apologise about, and I don't feel like our time has been wasted; rightly or wrongly, this is a popular topic and some people get tired of explaining it to multiple people. Everyone starts out ignorant, but the one who dares to be a fool is the one who learns. $\endgroup$ – Calvin Khor Oct 6 at 9:25
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    $\begingroup$ You may find information on the Riemann zeta function slightly interesting. If you consider complex numbers, then we have $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ for $s=\sigma+it$ and $\sigma>1$, $t\in\Bbb R$. Riemann extended this to the complex plane, and it is true that $\zeta(-1)=-\frac{1}{12}$. This is one reason for the popular notion that $\sum n=-\frac{1}{12}$. $\endgroup$ – Clayton Oct 9 at 14:59
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    $\begingroup$ Possible duplicate of How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization $\endgroup$ – Xander Henderson Oct 13 at 17:18
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    $\begingroup$ This is not at all a duplicate of the linked post Xander gave. That one specifically mentions "zeta regularization" while OP in this one has yet to know what a divergent series is. There is a big difference between the contexts between these two posts. $\endgroup$ – Jack Oct 13 at 21:54
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Take two infinite sets: $A=\{1, 2, 3, \cdots\}$ and $B=\{4, 5, 6, \cdots\}.$ Then $A-B = \{1,2,3\}$ and so is of cardinality 3. So infinity minus infinity must be 3. Now by choosing a different $B$ you can make infinity minus infinity any number you like.

In fact, choose $B=\{1, 3, 5, \cdots\}$ and then $A-B =\{2, 4, 6, \cdots\},$ and so $A-B$ has cardinality equal to infinity.

The point here is that when you start juggling infinities, you can make just about anything equal to just about anything. Which means almost everything is not well-defined. Once you had a divergent series in your hand, everything else was nonsense.

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    $\begingroup$ Sorry for the inconvenience I was uninformed, i watch a Numberfile video that followed this line of reason. $\endgroup$ – Aops Vol. 2 Oct 5 at 1:14
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    $\begingroup$ $1+2+3+..=\infty$; the other relation means that there is a function $f$ which is originally defined only for numbers bigger than $1$ and then there is a clever way to extend that function to the whole real line in a super nice way and it so happens that $f(-1)=-\frac{1}{12}$. But now if you would naively plug $-1$ in the original definition of $f$ (not allowed of course strictly speaking but as they do it in physics, this is called regularization), you would get $1+2+...$, so one can say that in a clear definite sense we can regularize the infinity given by $1+2+..$ to be $-\frac{1}{12}$ $\endgroup$ – Conrad Oct 5 at 1:32
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    $\begingroup$ Or better, subtraction of infinities becomes nebulous. Addition of infinities is not. $\endgroup$ – The_Sympathizer Oct 5 at 22:57
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    $\begingroup$ -1. Divergent series can make perfect sense and be perfectly well-defined. Also, what in the world does this cardinality argument have to do with infinite series? $\endgroup$ – user76284 Oct 11 at 17:12
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    $\begingroup$ @Conrad That’s not the only way to obtain $\sum_{n=1}^\infty n = -1/12$. $\endgroup$ – user76284 Oct 11 at 17:15
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First of all, you have to understand the following point:

There is no single canonical way of defining 'summing infinitely many things'.

As a result, there are several different notions of infinite sums, some of which are not even compatible to each other. For instance, the followings are some selected summability methods:

\begin{align*} \begin{array}{|c|c|c|} \hline & \begin{array}{c}\textbf{Definition}\\ \scriptsize\text{(some notations are not standard)} \end{array} & \textbf{Examples} \\ % \hline \begin{array}{c} \text{ordinary}\\ \text{summation} \end{array} % & \displaystyle\sum_{n=1}^{\infty} a_n := \lim_{N\to\infty} \sum_{n=1}^{N} a_n % & \begin{array}{c} \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \\ \displaystyle\sum_{n=1}^{\infty} (-1)^n = \mathtt{undefined} \\ \displaystyle\sum_{n=1}^{\infty} 1 = \infty \end{array} \\ % \hline \begin{array}{c} \text{Abel}\\ \text{summation} \end{array} % & \displaystyle\sum_{n=1}^{\infty} a_n = \lim_{x\to1^-} \sum_{n=1}^{\infty} a_n x^n \quad \text{(A)} % & \begin{array}{cr} \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} & \text{(A)} \\ \displaystyle\sum_{n=1}^{\infty} (-1)^n = -\frac{1}{2} & \text{(A)} \\ \displaystyle\sum_{n=1}^{\infty} 1 = \infty & \text{(A)} \end{array} \\ % \hline \begin{array}{c} \text{Dirichlet} \\ \text{regularization} \end{array} % & \begin{array}{c} \displaystyle\sum_{n=1}^{\infty} a_n = \lim_{s\to0} D(a, s) \quad \text{(D)} \\ \scriptsize\text{where $D(a, s)$ is the Dirichlet series for $a = (a_n)_{n\geq 1}$} \end{array} % & \begin{array}{cr} \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} & \text{(D)} \\ \displaystyle\sum_{n=1}^{\infty} (-1)^n = -\frac{1}{2} & \text{(D)} \\ \displaystyle\sum_{n=1}^{\infty} 1 = -\frac{1}{2} & \text{(D)} \end{array} \\ % \hline \end{array} \end{align*}

Considering the plethora of different definitions, you must be very careful about which one you use. And usually in mathematics, any infinite summation is considered to be an ordinary one unless stated otherwise. So, make sure to explicitly state the method used in case it is different from the ordinary one.

Now let us take closer look on OP's question. Given the context, I suspect that OP is working under the Dirichlet regularization or Ramanujan summation. They are simply some systematic ways of assigning values to the symbol $\sum_{n=1}^{\infty} a_n$, whose exact detail requires advanced level of math. So let me sweep this under the rug, although we should remark one of the consequence of the definition:

Some summability methods do not necessarily follow all the familiar rules that hold for ordinary summation.

To emphasize this distinction, let's estrange us from the confusing old notation $a_1 + a_2 + a_3 + \cdots$ and instead adopt the following idiosyncratic one

$$ \mathtt{Sum?}[ a_1, a_2, a_3, \cdots ] \quad \text{or} \quad \mathtt{Sum?}[(a_n)_{n=1}^{\infty}] $$

for the summability method involved in OP. This will make easier to track the manipulation involved in and spot the flow in OP's computation. Also, we will assume that OP's summability method $\mathtt{Sum?}$ satisfies the following two properties

  • $\mathtt{Sum?}$ is linear. In other words,

    $$\mathtt{Sum?}[\alpha a + \beta b] = \alpha \, \mathtt{Sum?}[a] + \beta \, \mathtt{Sum?}[b] $$

    holds for any constants $\alpha, \beta$ and sequences $a = (a_n)_{n=1}^{\infty}$, $b = (b_n)_{n=1}^{\infty}$ which are summable under $\mathtt{Sum?}$.

  • $\mathtt{Sum?}[1,2,3,\cdots] = -\frac{1}{12}$, such as in Dirichlet regularization or Ramanujan summation.

The first one is satisfied by essentially any interesting summability methods, so we include this to our one as well. The second one was OP's starting point.

  1. OP's first step follows from linearity:

    $$ \mathtt{Sum?}[2,4,6,\cdots] = 2 \, \mathtt{Sum?}[1,2,3,\cdots] = -\frac{1}{6}. $$

  2. The second step alwo follows from linearity:

    \begin{align*} \mathtt{Sum?}[1,3,5,\cdots] &= \mathtt{Sum?}[2,4,6,\cdots] - \mathtt{Sum?}[1,1,1,\cdots] \\ &= -\frac{1}{6} - \mathtt{Sum?}[1,1,1,\cdots]. \end{align*}

  3. The next step is the problematic. What is attempted in this step can be rephrased as

    \begin{align*} \mathtt{Sum?}[1,3,5,\cdots] &\stackrel{?}= \mathtt{Sum?}[1, 0, 3, 0, 5, 0, \cdots] \\ &= \mathtt{Sum?}[1, 2, 3, 4, 5, 6, \cdots] - \mathtt{Sum?}[0, 2, 0, 4, 0, 6, \cdots] \\ &\stackrel{?}= \mathtt{Sum?}[1, 2, 3, 4, 5, 6, \cdots] - \mathtt{Sum?}[2, 4, 6, \cdots] \end{align*}

The issue is, we do not know whether inserting zeros to the sequence preserves the value of $\mathtt{Sum?}$ or not. Indeed, under the Dirichlet regularization, we have

\begin{align*} &\left[ \text{Dirichlet reg. of $1+3+5+\cdots$} \right] \\ &= \lim_{s \to 0} \left[ \text{analytic continuation of } s \mapsto \sum_{n=1}^{\infty} \frac{2n-1}{n^s} \right] \\ &= \lim_{s \to 0} (2\zeta(s-1) - \zeta(s)) = \frac{1}{3}, \end{align*}

while we get

\begin{align*} &\left[ \text{Dirichlet reg. of $1+0+3+0+5+\cdots$} \right] \\ &= \lim_{s \to 0} \left[ \text{analytic continuation of } s \mapsto \sum_{n=1}^{\infty} \frac{2n-1}{(2n-1)^s} \right] \\ &= \lim_{s \to 0} (1 - 2^{1-s}) \zeta(s) = \frac{1}{12}. \end{align*}

So, we cannot expect that $\mathtt{Sum?}[1,3,5,\cdots]$ and $\mathtt{Sum?}[1, 0, 3, 0, 5, 0, \cdots]$ have the same value.

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In mathematics, definitions are Most Important, and in the ordinary definition of $a_1+a_2+\dots=s$, every single sum you talk about diverges.

Rearrangement of terms is not allowed for the sense of summation that is needed to talk about your sums. Rearrangement is dangerous even when we are talking about the usual definition of convergence. For instance, consider

$$ 1-\frac12 + \frac13 - \frac14 +\dots = \log 2 $$ The value $\log 2$ is true under the ordinary definition of summation-

$a_1+a_2+\dots=s$ if for every real number $\epsilon>0$, there exists a natural number $N>0$ such that for all natural numbers $n>N$,$$ |a_1 + a_2 + \dots + a_n - s | < \epsilon $$

But you can infact rearrange this series to converge to any real number- this is the Riemann Rearrangement Theorem, or the Riemann Series Theorem.

The definition that makes $1+2+3+\dots = -1/12$ true is IMO too complicated to explain here (but others are trying, so give them a read), and comes from something from the field of Complex Analysis, called analytic continuation. Suffice to say, rearranging it by splitting it into even and odd parts, or combining $-(2+4+6+\dots)$ with every second term of a series...is probably wrong, and if correct, would require some serious justification.

In my opinion, you should "learn to walk before you run" by reading more about the normal definition of convergence of sequences and series, and then doing many exercises to check your understanding. Its been a while since I looked at a book for these things, but you could try Spivak's Calculus or these set of notes by Hunter. Other people who know of better resources should comment...

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    $\begingroup$ +1 for taking into account OP's background. $\endgroup$ – Jack Oct 5 at 14:35
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    $\begingroup$ I have voted to reopen the post because I do not consider it a duplicate of the lined one: I don't think OP even knows what an infinite series means, let alone "zeta regularization" that is mentioned in the linked post. $\endgroup$ – Jack Oct 5 at 14:40
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    $\begingroup$ @XanderHenderson I'm not Jack but 2¢ - the Zeta regularisation is the easiest meaningful answer to this question that I know of, so purposely avoiding it makes the question much more complicated. IMO any (meaningful, of which I'm not sure there is one now) answer to that question is as useful to this OP as if OP asked for the value of $1+\frac12+\frac14+\dots $ and I replied with all sorts of weird summations that are consistent with the standard summation, when really the OP should be introduced to the concept of a limit, etc. PS maybe its just me but I'd DV this answer on that question $\endgroup$ – Calvin Khor Oct 14 at 1:19
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    $\begingroup$ @XanderHenderson I think you are disagreeing with me, I'm saying that (A) that question is a significantly harder question, and (B) doesn't have any indisputably good answers, therefore (C) should not be closed as a duplicate. In fact I would say that you have it in reverse; the answers there are so off the mark that they're the ones that should be modified to be here, and someone like reuns(?) who has written something about actually different summation methods (I have seen one about exponential regulators but cannot recall where) could write an actual answer to that question. $\endgroup$ – Calvin Khor Oct 14 at 2:43
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    $\begingroup$ @XanderHenderson I say that the answers aren't good on the basis of knowing some of the theory, and with respect/decency/courtesy/hopefully no CoC violation/etc I think the people who wrote the answers there, and those who upvoted don't know it as well. If I should accept the number of votes as the more significant indication of a good answer, then I am more inclined to agree with closing as a duplicate $\endgroup$ – Calvin Khor Oct 14 at 2:48
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No, it isn't true, at least not with the standard definition of infinite summation.

The sum

$$\sum_{n=1}^{\infty} 1$$

is either undefined, or equals $\infty$. "$\infty$" is an element in the "extended real number system", not a real number, and has the property that

$$\infty - \infty$$

is undefined. Both your $S_\mathrm{odd}$ and $S_\mathrm{even}$ also sum to $\infty$, hence your subtraction of them invokes the above form and is undefined. Conversely, you can consider this as a proof that such a subtraction has to be undefined, else it will lead to inconsistencies in the number system - in particular, your last equality is

$$\infty = -\frac{3}{12}$$

which is nonsense of the same form as

$$2 = 1$$

and your "proof" is analogous to the the phenomenon which occurs in "fake proofs" of the above involving the similarly-undefined division by $0$.

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  • $\begingroup$ sorry for wasting your time, im 15 so i was very ignorant watching the Numberphile video. $\endgroup$ – Aops Vol. 2 Oct 5 at 22:58
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    $\begingroup$ @Aops Vol. 2 : no problem, I like answering questions on this site :) #mehhr. $\endgroup$ – The_Sympathizer Oct 5 at 22:59
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    $\begingroup$ Also, Numberphile's video has been criticized for this reason. It can be made legitimate, but only under the caveat that you are in effect using a non-standard, generalized concept of summation of an infinite series that assigns finite "sums" to series where the standard one would not, and the problem with it is that it did not emphasize this point strongly enough. $\endgroup$ – The_Sympathizer Oct 5 at 23:00
  • $\begingroup$ Ill have to be more careful :) thanks for explaining simply. Question: why is it that we can sum infinite geometric series? They both have infinite number of terms $\endgroup$ – Aops Vol. 2 Oct 5 at 23:04
  • $\begingroup$ There's a proof in a math book that also follows the reasoning i did above. $\endgroup$ – Aops Vol. 2 Oct 5 at 23:05

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