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

[Albert Chan]

(10-27-2024 12:36 PM)AnnoyedOne Wrote:  "1 + 2 + 3 +... = -1/12"
Really? Is that so? And you can "prove" it? Actually I'm pretty sure that you can't.

We just ASSIGN a finite value to this divergent sum.
Weird sum will get cancelled at the end of calculations.

We could even leave it un-evaluated as zeta(-1), but a number as place-holder is simpler.

F(n) = sum(k, k=n, inf) = zeta(-1) - sum(k, k=1..n-1) = zeta(-1) - n*(n-1)/2

sum(k, k=10..25) = F(10) - F(26) = (zeta(-1)-10*9/2) - (zeta(-1)-26*25/2) = 280

(08-19-2019 04:15 PM)Albert Chan Wrote:  Let \(t = \frac{a}{1-a}\), then \(T= \sum_{x=1}^{\infty} a^x u_x = t \{1 + (tΔ) + (tΔ)^2 + \cdots\}u_1 \)

We assign t = Σ(a^x, x=1..inf) = a/(1-a), even if |a|>1

Below example had a=2 --> t = 2/(1-2) = -2, which make no sense.
But, infinite placeholder will get cancelled, when we ask for finite sum. (*)

(08-19-2019 04:15 PM)Albert Chan Wrote:  

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

(*) 'infinite' formula was derived from finite sum, but a/(1-a) keep popping out.
It just happened that t = a/(1-a) = sum(a^x, x=1..inf)
Infinite placeholder did not really get cancelled, but it is OK

sum(a^x * u_x, x, m, n-1) = -a^(x-1) * t * {1 + (tΔ) + (tΔ)^2 + ...} u_x | x = m .. n

sum(2^x * x^2, x, 10, 26-1)
= preval(-2^(x-1) * t * (x^2 + (2x+1)*t + 2*t^2), 10, 26)
= 38788923392 - 67584
= 38788855808

---

More compact formula, for u in falling factorial form:

\(\displaystyle
\sum_{x=0}^{\infty} a^{x+1} \, x^\underline{n} = n! \, t^{n+1}
\quad \text{, where } t = \frac{a}{1-a}\)

\(\sum _{10}^{\infty} 2^x x^2
= 2^9 \sum _{0}^{\infty} 2^{x+1} (x+10)^2
= 2^9 \sum _{0}^{\infty} 2^{x+1} (x^\underline{2} + 21x + 10^2)
→ 2^9 \,t\, (2t^2 + 21t + 10^2)
= -67584\)

\(\sum _{26}^{\infty} 2^x x^2
= 2^{25} \sum _{0}^{\infty} 2^{x+1} (x+26)^2
= 2^{25} \sum _{0}^{\infty} 2^{x+1} (x^\underline{2} + 53x + 26^2)
→ 2^{25} \,t\, (2 t^2 + 53t + 26^2)
= -38788923392\)

Update equivalent formula, with r = 1/t = (1-a)/a = 1/a - 1

\(\displaystyle
\sum_{x=0}^{\infty} \frac{x^\underline{n}}{(r+1)^{x+1}}
= \frac{n!}{r^{n+1} }
\)