1+2+3+4+5+6+7+... = 3/16

If this post tells us that the sum of every natural number is 3/16, and blackpenredpen made a video about him proving that this sum is 1/8, AND Numberphile claimed that it's equal to -1/12, then I think that this sum has infinite results.

@Minh Flip Bottle well clearly it does. I can prove it is 0:

S = (1+3+5+7+9+...) + (2+4+6+8+10+...)

S = S(odd) + 2S

(1+3+5+7+9)+(11+13+15+17+19)+...

(5*5)+(5*15)+(5*25)+(5*35)+...

(5*5)(1+3+5+7+9+...)

25*S(odd) = S(odd)

24*S(odd) = 0

S(odd) = 0

S = 0 + 2S

S = 2S

S = 0

I can also generalize maxx's solution so it gives off whatever number you want

S = S = (1+3+5+7+9+...) + (2+4+6+8+10+...)

let n be an odd number (this will come in handy later)

(1+3+5+...+(2n-1)) + ((2n+1)+(2n+3)+(2n+5)+...)+... = S(odd)

(1+3+5+...+(2n-1)) + (2n+2n+2n+...+1+3+5+...)+(4n+4n+4n+...+1+3+5+...)+...

there are n (2n) terms (and 4n terms too but you get the point)

n^2+(2n^2+n^2)+(4n^2+n^2)+(6n^2+n^2)+...

2n^2(1+2+3+4+5+...)+(n^2+n^2+n^2+n^2+...)

2n^2*S + (n^2+n^2+n^2+n^2+...)

2n^2*2 + (n+n+n+n+...) + (n+n+n+n+...) + ...

S = 2n^2*S + ((n+n+n+n+...) + (2+4+6+8+10+...))

(n+n+n+n+n+...) + (2+4+6+8+10+...)

(2+n)+(4+n)+(6+n)+(8+n)+(10+n)+...

The sequence will start on n+2 (obviously)

since n is odd then the sequence is just S(odd) with a few missing terms

the missing terms are 1, 3, 5, ... until n

1+3+5+...+n = ((n+1)/2)^2

S = S(odd) + 2S // from earlier in the S = 0 proof

S(odd) = S - 2S

S(odd) = -S

(2+n)+(4+n)+(6+n)+(8+n)+(10+n)+... = S(odd) - 1+3+5+...+n

(2+n)+(4+n)+(6+n)+(8+n)+(10+n)+... = -S - ((n+1)/2)^2 // substitute

S = 2n^2*S - S - ((n+1)/2)^2

-(2n^2-2)*S = -((n+1)/2)^2 // move all the S terms to the left

(2n^2-2)*S = ((n+1)/2)^2

S = (n+1)^2/8(n^2-1)

S = (n+1)/8(n-1) // knock out a common (n+1) factor

boom. done (almost)

just to check, let's plug in 5 because that's what maxx used:

S = (5+1)/8*(5-1)

S = 6/8*4

S = 6/32

S = 3/16

note a couple things:

1. we only used the fact that n is odd because we needed a concrete way to express S(even) in terms of S(odd)

2. plugging in n = 1 (a.k.a. individual grouping) gives us undefined (which is what common sense dictates)

3. as n tends to infinity (a.k.a. grouping the entire thing), S tends to 1/8

4. plugging in n = 0 (a.k.a. no grouping) gives us -1/8

5. S is only negative when -1 < n < 1

6. to get out an integer result (or at least a result greater than 1) then 1 < n < 9/7

7. to pick out any S you want then n = (8S+1)/(8S-1)