Surreal numbers - does the triangle number formula work?

Question

In surreal numbers, is there a solution to: 1+2+3+...+ω? In particular, is the solution ω(ω+1)/2?

Intuitively, it seems like it should still work because I can imagine putting these two triangles together to form a rectangle that is almost a square (one side ω, one side ω+1).

Answer 1

To do this you need the transfer principle which is only available in the hyperreals (see nonstandard-analysis tag), not the surreals.

Answer 2

The problem lies with your sum, it is not necessarily well-defined, as the comments pointed out. Let's try to define it for fun!

Simple Approach

For $n\in\mathbb{N}$ and any surreal number $x$ we can do normal summation with finite summands like $$\sum_{k=1}^n (x-n)= nx-\frac{n(n+1)}{2}$$ (Proof by induction and commutative/associative law of surreal numbers.) Note that $x=\omega$ is well possible.

Now, if we set $$S^0_n := \sum_{k=1}^n k,\quad S^\infty_n := \sum_{k=1}^n (\omega-k), \quad \sum_{k=1}^{\omega} k := \omega+\lim_{n\rightarrow\infty} \left(S^0_n+S^\infty_n\right)$$ From surreal numbers, we know that $\{\omega,2\omega,3\omega,\ldots|\}=\omega^2$, so it is an acceptable (and rather arbitrary) assumption to have $\lim_{n\rightarrow\infty} n\omega = \omega^2$, therefore, in this case, we would have $$\sum_{k=1}^\omega k = \omega + \omega^2\neq\frac{\omega(\omega+1)}{2}$$

However

Assume we set $${S'}_{n}^{\infty} := \sum_{k=0}^n (\omega-k)$$ Then we would say $$\sum_{k=1}^\omega k = \lim_{n\rightarrow\infty} \left(S^0_n+{S'}^\infty_n\right) = \lim_{n\rightarrow\infty} \left((n+1)\omega\right) = \omega^2$$

As I already hinted, the reason for this is that the limit of $n\omega$ is not really that well-defined, also we can't just take an $\omega$ out of the sum and the way how we defined the sum is pretty arbitrary. More precisely, the series doesn't converges normally anymore. You don't notice this in real numbers, as both variants evaluate to $+\infty$. The structure of hyperreal numbers and its transfer principle indeed ensures that $$1+2+3+\ldots\omega = \frac{\omega(\omega+1)}{2}$$ as Mikhail already noted. As the hyperreal numbers can be inbedded in the surreal numbers, I see no harm to say that it holds in surreal numbers, too. But only if you don't change the order.