Question

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

Answer 1

A few thoughts...

Explanation:

Conventionally, the infinite sum `1+2+3+...` 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(s) = sum_(n=1)^oo 1/n^s`

Note that this formula only converges when the real part of `s` is greater than `1`, but `zeta(s)` 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(-1) = -1/12`.

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

`sum_(n=1)^oo 1/(n^(color(blue)(-1))) = sum_(n=1)^oo n = 1+2+3+4+...`

Hence we might reasonably associate the value `zeta(-1) = -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+... = -1/12`

His first notebook contained a derivation something like this:

Let `c = 1+2+3+4+5+...`

Then we have:

`color(white)(+1)c = 1 + 2 color(black)(+ 3) + 4 color(black)(+ 5) + color(white)(0)6 + ...`

`color(black)(-4)c = color(white)(0) - 4 color(white)(+ 0) - 8 color(white)(+ 0) -12 color(white)(+ 0) ...`

`color(black)(-3)c = 1 - 2 color(black)(+ 3) - 4 color(black)(+ 5)-color(white)(0)6 +... = 1/4`

`:. c = -1/12`

Where did he get `1-2+3-4+5-6+... = 1/4` from?

Note that:

`sum_(n=0)^oo 1/x^n = 1/(1-x)`

So:

`sum_(n=0)^oo (-n/x^(n+1)) = d/(dx) sum_(n=0)^oo 1/x^n = d/(dx) 1/(1-x) = 1/(1-x)^2`

This converges for `abs(x) < 1`.

Then putting `x=-1` we find:

`0+1-2+3-4+... = 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 `-1/12` for `1+2+3+4+...`