Is $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ true?

Question

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.

Answer 1

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.

Answer 2

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...

Answer 3

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.

Answer 4

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 $0$.