# Why is 1+2+3+... = -1/12 ?

Dec 25, 2017

A few thoughts...

#### Explanation:

Conventionally, the infinite sum $1 + 2 + 3 + \ldots$ does not converge, but there are several ways of assigning a finite value to it.

Riemann zeta function

The most conventional method uses analytic continuation of the Riemann zeta function:

$\zeta \left(s\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ s$

Note that this formula only converges when the real part of $s$ is greater than $1$, but $\zeta \left(s\right)$ is a well behaved function that can be analytically continued to give values for all other values of $s$ except $s = 1$.

In particular $\zeta \left(- 1\right) = - \frac{1}{12}$.

Note that if we put $s = - 1$ in the summation formula for $\zeta \left(s\right)$ then we get:

${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{\textcolor{b l u e}{- 1}}} = {\sum}_{n = 1}^{\infty} n = 1 + 2 + 3 + 4 + \ldots$

Hence we might reasonably associate the value $\zeta \left(- 1\right) = - \frac{1}{12}$ with this divergent sum.

Ramanujan summation

Ramanujan found a couple of other methods to associate finite values with divergent sums, the first of which gives the same result:

$1 + 2 + 3 + 4 + \ldots = - \frac{1}{12}$

His first notebook contained a derivation something like this:

Let $c = 1 + 2 + 3 + 4 + 5 + \ldots$

Then we have:

$\textcolor{w h i t e}{+ 1} c = 1 + 2 \textcolor{b l a c k}{+ 3} + 4 \textcolor{b l a c k}{+ 5} + \textcolor{w h i t e}{0} 6 + \ldots$

$\textcolor{b l a c k}{- 4} c = \textcolor{w h i t e}{0} - 4 \textcolor{w h i t e}{+ 0} - 8 \textcolor{w h i t e}{+ 0} - 12 \textcolor{w h i t e}{+ 0} \ldots$

$\textcolor{b l a c k}{- 3} c = 1 - 2 \textcolor{b l a c k}{+ 3} - 4 \textcolor{b l a c k}{+ 5} - \textcolor{w h i t e}{0} 6 + \ldots = \frac{1}{4}$

$\therefore c = - \frac{1}{12}$

Where did he get $1 - 2 + 3 - 4 + 5 - 6 + \ldots = \frac{1}{4}$ from?

Note that:

${\sum}_{n = 0}^{\infty} \frac{1}{x} ^ n = \frac{1}{1 - x}$

So:

${\sum}_{n = 0}^{\infty} \left(- \frac{n}{x} ^ \left(n + 1\right)\right) = \frac{d}{\mathrm{dx}} {\sum}_{n = 0}^{\infty} \frac{1}{x} ^ n = \frac{d}{\mathrm{dx}} \frac{1}{1 - x} = \frac{1}{1 - x} ^ 2$

This converges for $\left\mid x \right\mid < 1$.

Then putting $x = - 1$ we find:

$0 + 1 - 2 + 3 - 4 + \ldots = \frac{1}{4}$

So Ramanujan's summation seems not to have anything directly to do with analytically continuing the Riemann zeta function, but results in the same value $- \frac{1}{12}$ for $1 + 2 + 3 + 4 + \ldots$