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

##### 1 Answer

A few thoughts...

#### Explanation:

Conventionally, the infinite sum

**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

In particular

Note that if we put

#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

**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

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

Then putting

#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