Sum of all natural numbers is 0?

Solution 1:

The simple answer to what seems to be your question: "why is my method wrong?"

You are assuming that a clearly divergent series converges to some number $S$; this provides a contradiction, if you assume a contradiction to be true then you can essentially use it to prove whatever nonsense you want.

Showing that this series sums to $-1/12$ and using this result does not have much to do with what you are thinking of as the sum of an infinite series. The Numberphile video is entertaining, but it is trash as far as actual mathematics. They don't explain this result, the meaning behind it, the context in which it is used, or the justification for the steps they use in their "proof". There is nothing rigorous about the proof they present, it's all smoke and mirrors. So it is natural that, by imitating their proof, you are ending up with a wrong result.

Solution 2:

The error in your derivation - I kid you not - is the assertion that 1+2=3 (and similarly 2+3=5, 3+4=7 etc.)

Sums like this get their meaning from Zeta function regularization, and the intuitive algebraic manipulations are only shorthand for correct manipulations on corresponding Dirichlet series. It so happens that the manipulations in the linked proof can be translated to proper ones, while your manipulations cannot.

In particular, it's easier to work with alternating sums because they can be thought of as formal power series in which you then substitute 1 (and since shifting is equivalent to multiplying by a power of $x$, you can do that freely). But you can't do that for sums like 1+2+3+... because you would get a 0 denominator, so other techniques are needed.

If your manipulations are translated to Zeta Function regularization, you'll get that the 1 and 2 you try to add are not really 1 and 2, they are $1^{-s}$ and $2^{-s}$, and these do not add up to $3^{-s}$, rendering the rest of the derivation invalid.

See also https://en.wikipedia.org/wiki/1_+2+3+4+_%E2%8B%AF#Heuristics.

Solution 3:

The most common way to define $\sum_{n\geq1}n$ is by its partial sums: $$ \sum_{n\geq1}n\equiv\lim_{N\rightarrow\infty}s_{N}\text{ where }s_{N}\equiv\sum_{n=1}^{N}n=\frac{1}{2}N\left(N+1\right). $$ Clearly, this is divergent.

As you seem to have noticed, there are other ways to define this sum. This causes all sorts of confusion, since people like to also use the notation $\sum_{n\geq1}n$ in these alternative definitions.

Let's now look at "zeta function regularization", which is the interpretation which yields $-1/12$. In particular, note that $$ \zeta(s)=\sum_{n\geq1}n^{-s}\text{ for }\operatorname{Re}(s)>1 $$ by definition. In the above, we are using the usual notion of summation by a limit of partial sums. The $\zeta$ function, however, is also defined for other values of $s$ by analytic continuation. We can thus reinterpret $$ \check{\sum_{n\geq1}}n\equiv\zeta(-1)=-1/12. $$ However, note that the two interpretations $$\sum\text{ and }\check{\sum}$$ are completely different!

As for the method which you refer to as "abusing divergent and oscillating series", perhaps you are thinking of this. This is not rigorous.