1 + 2 + 3 + ... = -1/12
08-19-2019, 04:15 PM (This post was last modified: 08-19-2019 07:47 PM by Albert Chan.)
Post: #15
 Albert Chan Senior Member Posts: 857 Joined: Jul 2018
RE: 1 + 2 + 3 + ... = -1/12
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 Ta - Tb still correct.
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 Messages In This Thread 1 + 2 + 3 + ... = -1/12 - Albert Chan - 10-01-2018, 08:26 PM RE: 1 + 2 + 3 + ... = -1/12 - Thomas Klemm - 10-01-2018, 09:09 PM RE: 1 + 2 + 3 + ... = -1/12 - SlideRule - 10-01-2018, 09:25 PM RE: 1 + 2 + 3 + ... = -1/12 - Albert Chan - 10-01-2018, 11:34 PM RE: 1 + 2 + 3 + ... = -1/12 - SlideRule - 10-01-2018, 11:56 PM RE: 1 + 2 + 3 + ... = -1/12 - Maximilian Hohmann - 10-02-2018, 05:52 PM RE: 1 + 2 + 3 + ... = -1/12 - pier4r - 10-03-2018, 03:42 PM RE: 1 + 2 + 3 + ... = -1/12 - Maximilian Hohmann - 10-03-2018, 04:15 PM RE: 1 + 2 + 3 + ... = -1/12 - pier4r - 10-03-2018, 05:13 PM RE: 1 + 2 + 3 + ... = -1/12 - Thomas Okken - 10-03-2018, 06:22 PM RE: 1 + 2 + 3 + ... = -1/12 - Maximilian Hohmann - 10-03-2018, 06:42 PM RE: 1 + 2 + 3 + ... = -1/12 - Albert Chan - 10-03-2018, 11:13 AM RE: 1 + 2 + 3 + ... = -1/12 - mfleming - 10-04-2018, 01:38 AM RE: 1 + 2 + 3 + ... = -1/12 - SlideRule - 10-04-2018, 11:48 AM RE: 1 + 2 + 3 + ... = -1/12 - Albert Chan - 08-19-2019 04:15 PM RE: 1 + 2 + 3 + ... = -1/12 - toml_12953 - 08-19-2019, 05:22 PM RE: 1 + 2 + 3 + ... = -1/12 - ijabbott - 08-19-2019, 07:17 PM RE: 1 + 2 + 3 + ... = -1/12 - jebem - 08-20-2019, 04:03 PM RE: 1 + 2 + 3 + ... = -1/12 - Gerson W. Barbosa - 08-20-2019, 11:31 PM

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