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.

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