# [FOM] Divergent series

joeshipman@aol.com joeshipman at aol.com
Sat Sep 22 09:47:18 EDT 2007

```What is 1+2+3+4+...?

Well, Riemann-zeta(s) = 1^(-s) + 2^(-s) + 3^(-s) + ...so we want
zeta(-1) which by analytic continuation is 1/12.

Here's another way to do it:
The function f(x)=(1-x)^(-2) has the power series expansion (1 + x +
x^2 + x^3 + ...)(1 + x + x^2 + x^3 + ...)=(1 + 2x + 3x^2 + 4x^3 + ...)

This series still doesn't converge at x=1, but not only can we
analytically continue it to get 1/12 as in the previous case, we can
get the same answer by algebra:

f(-1) = 1/4 = 1 - 2 + 3 - 4 + 5 - ...

Let S = 1+ 2 + 3 + 4 + ... and T = 1 - 2 + 3 - 4 + ...

Then
S+T = 2(1 + 3 + 5 + ...)
2S = 2 + 4 + 6 + ...
so
((S+T)/2) + 2S = S

T = -3S
1/4 = -3S
S=1/12

I got the same result by 2 different "methods", but it's conceivable
that I could have gotten a different answer by using yet another
"method". My question is,

What is the most general way known to consistently assign sums to
divergent series a0 + a1 + a2 + ... ? "Analytic continuation of a0 +
a1x + a2x^2 + ...to x=1 when the answer is unique" obviously is
consistent, but are there other methods consistent with this which
extend to situations where the power series has a 0 radius of
convergence and so cannot be analytically continued?

-- JS
________________________________________________________________________
Email and AIM finally together. You've gotta check out free AOL Mail! -
http://mail.aol.com
```