## Sum of natural numbers

The idea of this post is to highlight a remarkable result: The sum of all the natural numbers to infinity is:

$\displaystyle S = \sum_{i=1}^\infty n = 1 + 2 + 3 + 4 + 5 + \dots + \infty = -\frac{1}{12}$.

First, we may consider a different sum, $S_1$, (to $\infty$):

$S_1 = 1 - 1 + 1 - 1 + 1 - 1 + 1 \dots$,

On first inspection, one might think the value of $S_1$ is $0$ (i.e and infinite sum of $(1-1)$‘s) or $1$ (i.e $1$ plus an infinite sum of $(-1+1)$‘s). However, it’s actual value can be determined algebraically. If we write out $1 - S_1$,

\begin{aligned} 1 - S_1 &= 1 - (1 - 1 + 1 - 1 + 1 - 1 + 1 \dots)\\ &= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 \dots = S_1 \end{aligned}

we find that

$1 - S_1 = S_1$.

This rearranges to,

$\displaystyle S_1 = \frac{1}{2}$,

which is sensible, it can be thought of as the average of ending the sum at a plus $1$ and at a minus $1$.

Secondly, we can consider another sum, $S_2$

$S_2 = 1 - 2 + 3 - 4 + 5 - 6 \dots$,

that, when added to itself (with the help of visual alignment) reduces to

\begin{aligned} 2S_2 &= 1 - 2 + 3 - 4 + 5 - 6 + \dots \\ &+ \underline{0 + 1 - 2 + 3 - 4 + 5 + \dots} \\ &= 1 - 1 + 1 - 1 + 1 - 1 \dots \end{aligned}.

Which is $S_1$, and gives rise to a new algebraic expression,

$\displaystyle 2S_2 = S_1 = \frac{1}{2}$,

that may be solved giving a value of $S_2$,

$\displaystyle S_2 = \frac{1}{4}$.

This result can be used to help us evaluate $S$, if we compute $S - S_2$,

\begin{aligned} S - S_2 &= 1 + 2 + 3 + 4 + 5 + 6 + \dots \\ &- \underline{1 - 2 + 3 - 4 + 5 - 6 \dots} \\ &= 0 + 4 + 0 + 8 + 0 + 12 + \dots \end{aligned}

and extract a factor of four from each term,

\begin{aligned} S - S_2 &= 4 + 8 + 12 \dots \\ &= 4(1 + 2 + 3 \dots) = 4S \end{aligned}

then rearranging what is left gives,

$\displaystyle S = -\frac{1}{12}$.

Thus, the sum of all the positive numbers unto and including infinity is not infinity, but $-\frac{1}{12}$.

Note that credit should be given to Numberfile for the videos (a) and (b), where i picked up on this.