1 + 2 + 3 + ... = -1/12

[Albert Chan]

I finally see the logic of assignment a divergent sum to a value.

Example: find \(\sum _{10}^{25} 2^x x^2 \) with divergent "sum"

t = \(\sum _{1}^{\infty} 2^x → \frac{2}{1-2} → -2 \)

Getting coefficients of polynomial of t, to calculate the "sum"

Code:

x  x^2 Forward Difference Table
10 100
11 121 21
12 144 23 2

26 676
27 729 53
28 784 55 2

\(\sum _{10}^{\infty} 2^x x^2 = 2^9 \sum _{1}^{\infty} 2^x (x+9)^2 → 2^9 (100t + 21t^2 +2 t^3) → -67584\)

\(\sum _{26}^{\infty} 2^x x^2 = 2^{25} \sum _{1}^{\infty} 2^x (x+25)^2 → 2^{25} (676t + 53t^2 +2 t^3) → -38788923392\)

\(\sum _{10}^{25} 2^x x^2 = \sum _{10}^{\infty} 2^x x^2 - \sum _{26}^{\infty} 2^x x^2 → 38788855808\)

Amazingly, it match true result


Update: t polynomial trick always work. It is based on summation formula:

\(F(x) = (\frac{-a^x}{1-a}) \{1 + (\frac{aΔ}{1-a}) +
(\frac{aΔ}{1-a})^2 + \cdots \} u_x \)

\(\sum_{x=0}^{n-1} a^x u_x = F(n) - F(0) \)

Above formula, being a finite sum, does not require convergence.

t polynomial trick assumed x from 1 to ∞, and F(∞) = 0, and simplify to T = -F(1):

Let \(t = \frac{a}{1-a}\), then \(T= \sum_{x=1}^{\infty} a^x u_x = t \{1 + (tΔ) + (tΔ)^2 + \cdots\}u_x \)

For divergent sum, assumption of F(∞) = 0 is wrong, but finite sum T_a - T_b still correct.