# $1+1+1+1+\cdots = -\frac{21}{24}$ [closed]

In this video of numberphile they show a series of calculations to show that the sum of natural numbers is -1/12. Then I tried to proof other things using these mathematics techniques and I found the following... is it correct? there are a specific rules when trating these numbers?

The sum of natural numbers is:

$S_1 = 1+2+3+4+5+\cdots = -\frac{1}{12}$

Another summatory:

$S_2 = 1-1+1-1+\cdots = \frac12$ (proof)

And I want to calculate $S_3$:

$S_3 = 1+1+1+1+\cdots$

Lets add $S_1$ and $S_2$:

\begin{align} S_1=&1+2+3+4+5+6+7+8+9+10+\cdots\\ S_2=&0+1+0-1+0+1+0-1+0+1+\cdots\\ S_1+S_2 =& 1+3+3+3+5+7+7+7+9+11+\cdots\\ S_1+S_2 =& 1+5+9+13+\cdots\\ &3(3+7+11+15+\cdots) \end{align}

Adding $2S_3$ to this sum: \begin{align} S_1+S_2+2S_3 =& 1+5+9+13+\cdots+2+2+2+2+\cdots\\ &3(3+7+11+15+\cdots)\\ S_1+S_2+2S_3 =& 3+7+11+15+\cdots\\ &3(3+7+11+15+\cdots)\\ S_1+S_2+2S_3 =& 4(3+7+11+15+\cdots)\qquad (Eq.1) \end{align}

On the other hand: \begin{align} S_1 &= 1+2+3+4+5+\cdots\\ S_1 &= 0+1+2+3+4+\cdots\\ 2S_1 &= 1+3+5+7+9+\cdots \end{align}

And also

\begin{align} 2S_1 &= 2(1+2+3+4+5+\cdots)\\ 2S_1 &= 2+4+6+8+10+\cdots \end{align}

Then if we add these results

\begin{align} 2S_1 &= 1+3+5+7+9+\cdots\\ 2S_1 &= 2+4+6+8+10+\cdots\\ 4S_1 &= 3+7+11+15+19+\cdots \end{align}

Replacing in Eq.1 we have

\begin{align} S_1+S_2+2S_3 &= 4(4S_1)\\ S_3 &= \frac{15S_1-S_2}2\\ S_3 &= \left(-15\left(\frac1{12}\right)-\frac12\right)\frac12\\ S_3 &= -\frac{21}{24}\\ 1+1+1+1+\cdots &= -\frac{21}{24} \end{align}

However this sum corresponds with Riemann zeta function for $s=0$, which is $\zeta(0)=-\frac12$.

My question is, there is something wrong? there are specific rules when trating with these summatories?

## closed as unclear what you're asking by Peter Franek, Did, user91500, Parcly Taxel, WatsonNov 2 '16 at 12:50

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• If you assign a finite value $A$ to $1+1+1+\ldots$, then $A+1 = 1+(1+1+1+\ldots) =1+1+1+\ldots = A$ thus it is inconsistent and from this you can get the value you want – reuns Nov 1 '16 at 7:26
• Now if you want something consistent, use an analytic continuation method, or apply the rules (and only the rules) corresponding to one of those (see this question) – reuns Nov 1 '16 at 7:35
• Anything can be derived if you start with a bunch of nonsense with unclear meaning. – Peter Franek Nov 1 '16 at 7:44
• Can't someone just delete that video from the Internet, please... – Hans Lundmark Nov 1 '16 at 9:41
• The video you link to is a shame (an annoying (to me) feature of the situation being that this is partly financed by bona fide mathematical professional associations...). Actually, later on, a second video by the same team appeared, as part of a damage control operation, relying on typical "I was caught red handed saying crap but let me continue to pretend otherwise" techniques. For a decent take on the whole $-1/12$ thing, and to actually learn some maths, one could rather watch Ramanujan: Making sense of 1+2+3+... = -1/12 and Co. by Mathologer. – Did Nov 1 '16 at 12:15

I want to remind you that there is no genuine arithmetic operation like 'summing infinitely many numbers'. We just try to be as close to this nebulous intuition as possible. Depending on which aspect you want to reflect on your theory, however, you sometimes end up with mutually incompatible definitions. So it is extremely important to know in which notion of infinite summation you are working with.

In your question, you are working under the notion of zeta function regularization. (Let me use the notation $\sum^{\mathfrak{R}}$ for zeta-regularized sums to distinguish them from usual sums.) This summation has linearity: if $\sum^{\mathfrak{R}} a_n$ and $\sum^{\mathfrak{R}} b_n$ exist, then

$$\sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} (a_n + b_n) = \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} a_n + \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} b_n \qquad \text{and} \qquad \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} c a_n = c \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} a_n.$$

On the other hand, this summation lacks stability in the sense that

$$\sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} a_n \qquad\text{and}\qquad a_1 + \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} a_{n+1}$$

are often not equal! For instance,

\begin{align*} 1+1+1+\cdots\text{''} &=\sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} 1 = \zeta(0) = -\frac{1}{2},\\ 1 + (1+1+1+\cdots)\text{''} &=1 + \bigg( \sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} 1 \bigg) = 1 + \zeta(0) = \frac{1}{2}. \end{align*}

and similarly

\begin{align*} 1+2+3+\cdots\text{''} &=\sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} n = \zeta(-1) = -\frac{1}{12},\\ 0+1+2+\cdots\text{''} &=\sum_{n=1}^{\infty} {\vphantom{\Big)}}^{\mathfrak{R}} (n-1) = \zeta(1) - \zeta(0) = \frac{5}{12}. \end{align*}

This invalidated your argument since it heavily relies on this forbidden manipulation.

• Genius! Never thought about using quotations, and they work perfectly here. – Simply Beautiful Art Nov 1 '16 at 16:45

Here is a similar manipulation:

$$S=1-\frac12+\frac13-\frac14+\dots$$

Now rearrange it so that it follows an odd even even pattern.

$$=1-\frac12-\frac14+\frac13-\dots$$

In every three terms, add the first two and leave the third.

$$=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)+\dots$$

$$=\frac12-\frac14+\frac16-\dots$$

$$=\frac12S$$

Which is strange because common knowledge says $S=\ln(2)$, so we basically proved that $\ln(2)=\frac12\ln(2)$, which is clearly false.

When dealing with such manipulations, you must be careful, especially when dealing with divergent series.