Sum of convergence series in non-standard analysis

Question

We know $\sum_{n=1}^\infty \frac{1}{2^n} = 1$. Here’s an informal proof:

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +... = S$

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +... = 2S$

But this is just $1 + S = 2S$, so $S=1$.

Will this change if we consider infinitesimals? That is, if we allow for the existence of infinitesimals, is $\sum_{n=1}^\infty \frac{1}{2^n} = 1$?

Edit: by “allow for the existence of infintesimals”, I just mean in nonstandard analysis. Is $\sum_{n=1}^\infty \frac{1}{2^n} = 1$ there still?

What do you mean by "allow for the existence of infinitesimals", precisely? — Eric Wofsey, Dec 16 '19 at 05:39
You are heading for a binary version of Is $0.99999\cdots = 1$? — Henry, Dec 16 '19 at 05:42
I don’t know how to phrase it. I take it that $\sum_{n=1}^\infty \frac{1}{2^n} = 1$ in real analysis, not nonstandard analysis. Just wondering if this will change in nonstandard analysis. — phst, Dec 16 '19 at 05:42

score 1 · Accepted Answer · answered Dec 16 '19 at 07:53

Nothing about the arithmetic of standard reals changes when you allow nonstandard reals to exist (in the sense of Robinson's ultrapower construction).

I think you might instead be asking: "how does infinite summation work in nonstandard analysis"?

In that case, you are asking the equivalent question: is $\mathrm{st}\left(\sum_{n=1}^N \frac{1}{2^n} \right)$ well-defined, and what is its value, when $N$ is any nonstandard (infinite) natural. The answer, of course, is $1$. Indeed, $$1-\sum_{n=1}^m \frac{1}{2^n} = 1 - \frac{1}{2^m}$$ when considered in the standard world, and this is a statement that transfers to the hyperreals, so it is the case that $$1-\sum_{n=1}^N \frac{1}{2^n} = 1-\frac{1}{2^N}$$ when $N$ is an infinite hypernatural. But that number is only infinitesimally far from $1$.

Sum of convergence series in non-standard analysis

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