# Infinite sum of ones

Consider the following infinite series:

M = 1 + 1 + 1 + 1 + 1 + 1 + 1 + ... = $\zeta$(0) = -$\mathbf{\frac{1}{2}}$

Rewrite it that way:

M = (1) + (1 + 1) + (1 + 1 + 1) + (1 + 1 + 1 + 1) + ... = 1 + 2 + 3 + 4 + ...

Given that the sum of integers is proven to equal -$\mathbf{\frac{1}{12}}$, I obtain:

M = -$\frac{1}{12}$ $\Rightarrow$ -$\frac{1}{2}$ = -$\frac{1}{12}$

Which contradicts the initial statement. Where did I do wrong?

• There are numerous, numerous posts about this misconception that you can manipulate divergent series like this. Please do a search. If you watched that infamous Numberphile video, please note that I think it's done a great disservice to many. – Deepak Oct 2 '16 at 11:19
• have you searched sum of all natural number ? – Aster rovels Oct 2 '16 at 11:22
• I understand but my question is about the meaning of the value assigned to a divergent series. @Deepak – Matteo Pariset Oct 2 '16 at 11:23
• There's actually a lot of mathematical deepness within the theory of divergent series. This particular example has to deal with Analytic continuation in complex analysis. There's no simple explanation other than: one must be very careful about manipulating divergent series and giving a meaning to what $\sum$ might signifies in the case of divergent series. – Hermès Oct 2 '16 at 11:26
• @MatteoPariset There are ways to assign finite values to some divergent sums by techniques such as zeta function regularisation, Cesaro summation and Ramanujan summation. These may have some rigorous deep meaning but should not be treated as true sums of these series (no actual sum exists). Manipulations like grouping terms may not have the predicted effect on the "sum". – Deepak Oct 2 '16 at 11:35

So... this is not actually how we evaluate divergent series, and, note, that when talking about your particular sums, we say "zeta regularization", which is basically what you attempted to do.

So to start, I want to explain a bit of zeta regularization. In essence, it is the following:

$$\sum_{n=1}^\infty n^z=\zeta(-z)$$

The above is, on its own, not a true statement. First of all, see quite clearly that it makes no sense.

But then, you might wonder that if it makes no sense, how does $\zeta(z)$ acquire values for negative numbers in the first place?! Well, the whole idea stems from analytic continuation, as we usually define the Riemann zeta function as follows:

Let there be a meromorphic function (known as the Riemann zeta function) to be defined as follows,

$$\zeta(z)=\sum_{n=1}^\infty n^{-z}\tag{\boxed{\Re(z)>1}}$$

What all the above basically means is that we define the Riemann zeta function by that sum only when it makes sense. Secondly, we let it be analytically extendable to values for which the original definition does not make sense.

Now, analytic continuation is quite a magical thing, and I indeed hope you will one day understand it better. In essence, analytic continuation works for analytic functions, and allows such functions to extend beyond their normal domains.

One key aspect to the idea of analytic continuation is that regardless of how you do it, the result is the same. This is why, when you read the Wikipedia on the Riemann zeta function, you will find numerous forms of the function that all equal one another.

So in essence, zeta regularization is incorrectly using the original definition of the zeta function to define a divergent series.

But note that zeta regularization is not the only method you can use. We can regularize through other methods too.

I'd like to take a look at the following series, as it is easier to do things with these.

$$1-2+3-4+\dots$$

I'd like to regularize it as follows:

$$\lim_{r\to-1^+}\sum_{n=1}^\infty nr^{n+1}$$

The above is a perfectly valid method to regularizing a sum.

However,

$$1+1+1+1+\dots$$

can be made to equal anything. See clearly that

$$\sum_{n=1}^\infty1\ne\sum_{n=500}^\infty1$$

And yet, when you write them out, they appear to be the same.

The problem is that the order indeed matters. While $1+2=2+1$, moving terms around or adding them up differently results in problems.

A most famous case is displayed in an episode of the Simpsons.

The steps are indeed like what you attempted, just more hidden.

Another famous theorem relating here is the Riemann series theorem. It states that a conditionally convergent series can be rearranged to equal anything.

For example,

$$1-\frac12+\frac13-\frac14+\dots=1-\frac12-\frac14+\frac13-\frac16-\frac18+\dots+\frac1{2k-1}-\frac1{2(2k-1)}-\frac1{4k}+\dots$$

$$=\frac12-\frac14+\frac16-\frac18+\dots$$

$$1-\frac12+\frac13-\frac14+\dots=\frac12\left(1-\frac12+\frac13-\frac14+\dots\right)$$

Which is clearly a false statement.