Let S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + ...

If we isolate the odd and even numbers, we obtain

S = (1 + 3 + 5 + 7 + 9 + 11 + ...) + (2 + 4 + 6 + 8 + 10 + 12 + ...)

Let's start with the odd terms. Collect the terms in groups of 5 consecutive odd numbers:

(1+3+5+7+9) + (11+13+15+17+19) + (21+23+25+27+29) + ...

= 25 + 75 + 125 + 175 + 225 + 275 + ...

= 5*5 + (50 + 5*5) + (100 + 5*5) + (150 + 5*5) + ...

We can separate all the multiples of 50:

= (50 + 100 + 150 + 200 + 250 + ...) + (5*5 + 5*5 + 5*5 + 5*5 + 5*5 + ...)

= 50 (1+2+3+4+5+...) + (5*5 + 5*5 + 5*5 + 5*5 + 5*5 + ...)

= 50(S) + (5+5+5+5+5) + (5+5+5+5+5) + (5+5+5+5+5) + ...

Note: we obtain a single 5 for every odd term in the sum.

Now we still have the even terms remaining. Since we have an equal number of even terms and 5s from previously, we can add them together pair-wise:

(2 + 4 + 6 + 8 + 10 + ...) + (5 + 5 + 5 + 5 + 5 + ...)

= (2 + 5) + (4 + 5) + (6 + 5) + (8 + 5) + (10 + 5) + ...

= 7 + 9 + 11 + 13 + 15 + ...

Note: this sum is simply the sum of all odd numbers, minus the first 3 terms.

The sum of all odd numbers can be written as the sum of all numbers minus the sum of all even numbers:

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

Combined with the result above, we get 7 + 9 + 11 + 13 + 15 + ... = S - 2S - (1 + 3 + 5) = S - 2S - 9

Putting it together we have:

S = 50S + (S - 2S) - 9

S = 49 S - 9

-48 S = -9

S = 9/48 = 3/16

:)

what if not

∞

Well, Numberphile claimed that 1+2+3+4+5+…=-1/12.

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

oddnumber (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 oddthen the sequence is just S(odd) with a few missing termsthe 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:

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

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

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

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

S is only negative when -1 < n < 1

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

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

This kind of re-grouping of terms ONLY works if the original series is convergent, which in this case, it is clearly not. So you can probably group them in a myriad of ways to produce just about any sum you want. While that may somewhat amusing and/or entertaining it has no relevance whatsoever to assigning a finite sum to an infinite series. The assigning of S to the infinite sum in the first line is where the whole thing falls apart and everything after that is nonsense and just plain wrong.

Watch the Mathologer video on this and he will set you straight:

https://www.youtube.com/watch?v=YuIIjLr6vUASame holds for: 1-1+1-1+ ... = 1/2. This series diverges and the 1/2 is nonsense.