In the real number system, for example, the sum $\lim_{N \rightarrow \infty} \sum^N_{i=1} (\frac{1}{N}) = 1$, but the individual terms tend to zero due to the fact $\lim_{N \rightarrow \infty} \frac{1}{N} = 0$.
I naturally thought the hyperreal extension of the real numbers would be the next best place to look, but if my resource (and my deduction) is correct, it isn't.
The PDF at the bottom of the post states at section 3.2, "if ε and δ are infinitesimals, ε + δ is infinitesimal". This, to me, would mean that a sum of an infinite amount of hypperreals standard part (or shadow, as referenced in section 3.4 of the same PDF) would still be 0.
This goes against my intuition, however, considering that, according to section 6.1:
$$\lim_{x\rightarrow +\infty} f(x) = L \text{ iff } f(x) \simeq L \text{ for all } x \in *A^+_\infty$$
With $\simeq$ being defined as "infinitely close".
This implies to me that $\lim_{N \rightarrow \infty} \frac{1}{N} \simeq x $ for all $x \simeq 0$ and $x > 0$
Which then implies to me that since $\lim_{N \rightarrow \infty} \sum^N_{i=1} (\frac{1}{N}) = 1$, $\sum^H_{i=1} \delta \simeq 1$, with $\delta$ being an infinitesimal and with H being $\frac{1}{\delta}$?
But doesn't that violate the statement from section 3.2?
https://folk.uio.no/atodegaa/bachelor_project/hyperreals.pdf
Apologies if I sound a bit like a crank; merely someone quite new to nonstandard analysis.
Excuse the partial abuse of the {}^* notation, but I wanted to be as correct as possible.
$$(\forall x \in \mathbb{R}^+)(x \cdot \frac{1}{x} = 1) \rightarrow (\forall x \in {}^* \mathbb{R}^+)(x {}^*\cdot \frac{1}{x} {}^*= 1)$$ via transfer principal.
$$\forall x \in \mathbb{R}^+, x \cdot \frac{1}{x} = \underbrace{\frac{1}{x} + ... + \frac{1}{x}}_{x} = \sum^x_{i=1} \frac{1}{x} = \frac{x}{x} = 1$$
$$\forall H \in {}^* \mathbb{R}^+_\infty, H {}^*\cdot \frac{1}{H} {}^*= \underbrace{\frac{1}{H} {}^*+ ... {}^*+ \frac{1}{H}}_{H} {}^*= \sum^H_{i=1} {}^*\frac{1}{H} {}^*= {}^*\frac{H}{H} {}^*= 1$$
Good enough reasoning for me. If I'm incorrect, please leave a comment.