What is the sum of all complex integers?

Question

In line with

$$\zeta(-1)=-1/12$$ Could we, by considering

$$f(s)=\sum_{a,b\in\mathbb Z,\;(a,b)\neq(0,0)}\frac{1}{(a+bi)^{s}}$$ Evaluate the sum of all complex integers?

Answer 1

You might want to restrict to positive integers only since...

$$f(s) = \sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \sum_{a>0,b<0} \frac{1}{(a+ib)^s} + \sum_{a<0,b>0} \frac{1}{(a+ib)^s} \\+ \sum_{a<0,b<0} \frac{1}{(a+ib)^s} + \sum_{a=0,b\in Z}\frac{1}{(ib)^s} + \sum_{a\in Z,b=0} \frac{1}{a^s}$$

giving

$$f(s) = (1 + (-1)^s)\left[\left(\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + \frac{1}{(a-ib)^s}\right) + \zeta(s)(1+i^s)\right]$$

So $f(-1)$ (meaning the analytical continuation) will be $0$ (unless the analytical continuation of $\sum_{a>0,b>0} \frac{1}{(a+ib)^s} + ...$ has a simple pole at $s=-1$).

The same happens if one consideres the sum of all integers:

$$g(s) = \sum_{n\in Z, n\not= 0} \frac{1}{n^s} = (1 + (-1)^s)\sum_{n=1} \frac{1}{n^s} = \zeta(s)(1 + (-1)^s)$$

so the 'sum of all integers' are zero.

Answer 2

Necromancing this for another useful heuristic of a more elementary flavor.

Review:

We begin by recalling how to "derive" $1+2+3+4... = -\frac{1}{12}$

We observe that

$$ \sum_{n=0}^{\infty} e^{nx} = \frac{1}{1-e^x} \tag{1} $$

Via the geometric series formula $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

Now the following expansion can be considered:

$$ \sum_{n=0}^{\infty} e^{nx} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(nx)^{k}}{k!} \underbrace{=}_{\text{questionable step}} \ \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{(nx)^k}{k!} \tag{2} $$

Now we evaluate our expression:

$$ \sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{(nx)^k}{k!} = \sum_{k=0}^{\infty} \left[ \left( \sum_{n=0}^{\infty}n^k\right)\frac{x^k}{k!}\right] \tag{3} $$

So it's natural to equate the coefficient of the $k^{\text{th}}$ order term (multiplied by $k!$) in the laurent expansion of $\frac{1}{1-e^x}$ with $\sum_{n=0}^{\infty} n^k$ .

Recall:

$$ \frac{1}{1-e^x} = -\frac{1}{x} + \frac{1}{2} \mathbf{- \frac{1}{12}x} + ... \tag{4} $$

Therefore we conclude that $\sum_{n=0}^{\infty} n = -\frac{1}{12}$

Solution:

Having digested this we can now turn out attention to $\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} (n+mi)$

We first need our replacement for the geometric series $\sum_{n=0}^{\infty} x^n$ instead we want $\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} x^{n+mi}$

This has a closed form by expanding it as a 2dimensional grid and doing some algebra:

$$ \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} x^{n+mi} = \frac{1}{(1-x)(1-x^i)} \tag{5} $$

So now with extremely similar arguments involving expanding $e^{nx}$ as a summation over $k$ and re-arranging the $k$-sum to the front, we conclude that we want to extract the linear term from $$ \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} e^{(n+mi)x} = \frac{1}{(1-e^x)(1-e^{ix})} $$

We compute the laurent series::

$$ \frac{1}{(1-e^x)(1-e^{ix})} = -\frac{i}{x^2} - \frac{(1-i)}{2x} + \frac{1}{4} \mathbf{-\frac{1+i}{24}x} ... \tag{6} $$

Leading to the result @winther found using the zeta function.

Further Considerations:

Generally the sum $a+b\tau$ over non-negative integers $a,b$ is $-\frac{1+\tau}{24}$.

It is natural to consider then $3$-dimensional sums and higher of the form $a+b\tau + c\delta$ and these same techniques generalize there too.

One might be curious about structure which spans across dimension by consdering the sequence of numerators and denominators in $\sum n, \sum \sum n+m, \sum \sum \sum n+m+o ... $ given by $-\frac{1}{12}, -\frac{1}{12}, -\frac{19}{240} ... $. Unfortunately, this doesn't turn up any hits on OEIS. Oddly there seems to be some close connection between this series and these sequences. But it's not clear what the relationship is. Some kind of combinatorial or categorical interpretation here would probably clarify the matter significantly but that is well beyond my capabilities at this time :( .

Answer 3

I'm not sure how to go about proving it in any form of notation, but if we consider the real number axis from -x to +x, and the imaginary axis from -iy to +iy, then all the complex integers will be plotted as iy against x. Since there are no limits to the size of either real or imaginary parts and given that they may only be integers. I would have thought that the sum of all the complex integers would be zero. Mainly because the sum of all imaginary numbers from -iy to +iy is zero, and all the real numbers from -x to +x is zero. Am I missing something?

I see what you are saying, so 1+(1-1-1)+(1-1-1)...=-∞ But 1-(1-(1-1-1))-(1-(1-1-1))...=+∞

I'm not sure that eventually if all the combinations of number series is included in the argument a situation would arise where the net result is asymmetric.

Thank you Winther, I don't know if everyone will agree with you, but I do. But I'm beggining to see now that with complex numbers the problem may well be different.