Proof that sum of all positive integers is -1/12

Hi there,

I'm trying to use MATLAB to prove that the sum of all positive integers to infinity is -1/12 (sry but I don't know how to write in LaTeX). I've tried using a for loop as well as integrating, but the values I've attained from those methods just get larger and larger. I tried using Inf in my for loop but that led to the for loop iterating forever.

Here's what I have so far:

a = [];
for i = 1:10000
    if i ~= 10000
        a = [a, (i*(i+1))/2];
    else
        a = [a, (i*10000 + 1)/2];
    end
end
%disp(a)

plot(1:10000, a);

syms x
X = sum(a)
Y = -1*4*(10^-26)*x^5 - 1.6*(10^-22)*x^4 + 5*(10^-19)*x^3 + 0.5*x^2 + 0.5*x - 3*10^-11; %This I got from using Basic Fitting tool on MATLAB
Z = int(Y,1,1000) %Integration

How can I prove the proof using simple methods?

Thanks

[David Hill]

What, the sum of all positive integers to infinity? Is that not infinity?

[Jayesh Kamboj]

Intuition tells us that it should be infinity, but if you look at Ramanujan Sum (https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF) then it says that the sum should be equal to -1/12.

[Walter Roberson]

https://medium.com/cantors-paradise/the-ramanujan-summation-1-2-3-1-12-a8cc23dea793

Answer 1

[James Tursa]

This has nothing to do with MATLAB and numerical sums. You can pretty much get any answer you want by manipulating divergent series on paper, and manipulating divergent series doesn't prove anything ... and certainly is not going to match any numerical summing method you might do with MATLAB.

From the website:

     c = 1 + 2 + 3 + 4 + 5 + 6  + ... is a divergent series (inf)
    4c =     4     + 8     + 12 + ... is another divergent series (inf)
(c-4c) = 1 - 2 + 3 - 4 + 5 - 6  + ... is yet another divergent series (inf-inf on lhs)

The very first step is actually a violation of mathematical equality ... stating that the sum of a divergent series is equal to something. However, we can play the game for the first two steps. But for that last series, stating it is equal to c-4c is complete BS from a mathematical standpoint. You cannot manipulate divergent series this way and maintain equality in a mathematical sense. Similar manipulation can be done to get completely different results. And then "assigning" it a value of 1/(1+1)^2 by plugging in a value outside the domain of a convergent series for 1/(1+x)^2 because it looks similar is another bogus mathematical step.

This whole exercise is fun to play with, but does not have anything to do with mathematical equality.

EDIT

Here is my solution for that series sum.

Start with c and 2c and subtract them:

c = 1 + 2 + 3 + 4 + 5 + 6 + 7 + ...

-( 2c = 2 + 4 + 6 + ...)

c-2c = 1 + 3 + 5 + 7 + ...

So we get the result:

-c = 1 + 3 + 5 + 7 + ...

Multiply -c by 4 and rewrite it to get:

-4c = 4 + 12 + 20 + 28 + ...

-4c = 2 + 2 + 6 + 6 + 10 + 10 + 14 + 14 + ...

Now take that result and subtract our –c result:

-4c = 2 + 2 + 6 + 6 + 10 + 10 + 14 + 14 + ...

-( -c = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + ...)

-4c+c = 1 – 1 + 1 – 1 + 1 - 1 + 1 - 1 + ...

So clearly

-3c = 1 – 1 + 1 – 1 + 1 - 1 + 1 - 1 + ...

But we also have this series expansion

1/(1-x) = 1 + x + x^2 + x^3 + x^4 + ...

Plugging in x = -1 we get:

1/(1-(-1)) = 1 – 1 + 1 – 1 + 1 – 1 + ...

Combining the two results gives us the obvious answer:

-3c = 1/(1 – (-1))

-3c = 1/2

c = -1/6

So, I have proven that the -1/12 answer is clearly off by a factor of 1/2, right?

No, not really. Again, this is a fun exercise in manipulating divergent series, but I haven't followed the strict rules of mathematical equality.

[John D'Errico]

+1, even though this question is unresolvable using MATLAB, as James has recognized. Anyway, we know the answer to all questions must be 42, so he has clearly made a mistake.

[David Goodmanson]

Here's a take on the problem that does result in -1/12. Though the method is no more justifiable than James's, it uses a more straightforward process.

Consider the sum

1 + x + x^2 + x^3 + ...  = 1/(1-x)    

and its derivative

1 + 2x + 3x^2 + 4x^3 + ...  = 1/(1-x)^2    

Both of these series are convergent for |x| <1. For the first illegal step, plug in x = -1 and get

1 - 2 + 3 - 4 + 5 - 6 + ...  = 1/4    

Now consider

{A} = the set of all positive integers

{E} = the set of all even positive intergers

{O} = the set of all odd positive intergers

So the union of {E} and {O} is {A}, which is a true statement. Now denoting sum of elements of {A} = sA etc, we have

sA = sO + sE
sE = 2*sA
sO - sE = 1/4      as above

Solving these gives

sA = -1/12         the requred result
sE = -1/6
SO = 1/12

and as an added bonus we find that the sum of the odd positive integers = 1/12. Of course all three of the sum equations are bogus, but why not!

************************************* In a more serious vein:

The Riemann zeta function for a complex number s is defined as

zeta(s) = 1/1^s +1/2^s +1/3^s +1/4^s + ...   Re(s) > 1

but the partial sums of the series converge to a limit only for real(s) >1. The series is only convergent and meaningful for real(s) > 1. If you could plug in s = -1 you would have

zeta(-1) = 1 + 2 + 3 + 4 + ...

which is illegal because the series solution does not apply. The series is clearly divergent, besides which adding up a bunch of positive integers is not ever going to result in -1/12. But the Riemann zeta function can be analytically continued over to s = -1 in the complex plane. In that region zeta cannot be expressed in terms of the series above. But entirely different methods show that zeta(-1) = -1/12.

[Image Analyst]

There are tons of these kind of 1=0 (or whatever) "Proofs" on the internet: Click here

For example

1 + 1 = 1 + sqrt(1)

= 1 + sqrt(-1 * -1)

= 1 + sqrt(-1) * sqrt(-1)

= 1 + i * i

= 1 + -1

2 = 0

[Paul]

@David Goodmanson

Your recent edit reminded of this discussion that you may find to be of interest.

[David Goodmanson]

Hi Paul, it was interesting. At this point I hope that no one is contending that sym-vpa is anything other than wrong, when it sums 1/n or 1/sqrt(n) and comes up with a finite result.

[Paul]

I think your insights would be helpful over in that linked thread.