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?


Apologies if I sound a bit like a crank; merely someone quite new to nonstandard analysis.

  • $\begingroup$ Framing challenge: the series from your first line does not converge (to $1$ or anything else) because the harmonic series diverges. $\endgroup$ – Eric Towers Dec 2 '19 at 6:30
  • $\begingroup$ @EricTowers , maybe my notation is incorrect or I am misunderstanding you. That is meant to notate the sum of 1/N, N times, as N tends to infinity. For example, N=3 would be (1/3 + 1/3 + 1/3). If there is a better way to notate this (as I felt this was improper), I apologize. $\endgroup$ – Jarad Dec 2 '19 at 6:35
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    $\begingroup$ Perhaps "\sum_{i=1}^N \frac{1}{N}", giving "$\sum_{i=1}^N \frac{1}{N}$", which explicitly indicates you mean a finite sum, not an indefinite summation. $\endgroup$ – Eric Towers Dec 2 '19 at 6:37
  • $\begingroup$ @EricTowers went ahead and edited the post to hopefully clarify that. Thank you for correcting me :) $\endgroup$ – Jarad Dec 2 '19 at 6:42
  • $\begingroup$ Suppose $0<\delta_n<r$ for every standard real $r\in \Bbb R^+$ and each $n\in \Bbb N.$ Then for any standard $r\in \Bbb R^+$ and any $n,H\in \Bbb N, $ we have $\delta_n<r/H,$ so $\sum_{n=1}^H\delta_n<\sum_{n=1}^H(r/H)=r.$ So any standard $r\in \Bbb R^+$ is an upper bound for $\{\sum_{n=1}^H\delta_n: H\in \Bbb N\}.$ $\endgroup$ – DanielWainfleet Dec 2 '19 at 10:30

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.

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