# Convergence of 1/2^(1+infinitesimal)

If $a$ is an infinitesimal or a non-standard number or a number strictly larger then 0 and smaller then all positive real numbers. What is

$\sum_{n=1}^{\infty} \frac{1}{2^{1+a}}$

$\lim_{x\to \infty} (1+a)^x$

$\sum_{n=1}^{\infty} a$

-

Until kermit says what he means, we can only speculate.

Suppose $F$ is an ordered field, strictly larger than $\mathbb R$. Then there is an element $a \in F$ such that: $a>0$ but $a<r$ for all real $r>0$. Equivalently: $a>0$ but $a<1/n\;$ for all $n \in \mathbb N$. (Of course $\mathbb R$ contains a subset we naturally call $\mathbb N$.) I will consider a question in the same vein as kermit's questions, but--I think--more elementary:

$$\lim_{x\to\infty} xa$$

Since things like this are not conventionally defined in $F$, but only in $\mathbb R$, we will have to say what it means. There is more than one possibility. Here is one:

(1) Let $b \in F$. We say $\lim_{x\to \infty} xa = b$ iff for every positive $\epsilon \in F$, there exists $N \in \mathbb N$ such that for all $n \in \mathbb N$, if $n \ge N$, then $|na-b|<\epsilon$. If there is such an element $b$, then we say $\lim_{x\to\infty} xa$ exists.

(1') We say $\lim_{x\to \infty} xa = \infty$ iff for every $k \in F$, there exists $N \in \mathbb N$ such that for all $n \in \mathbb N$, if $n \ge N$, then $na > k$.

Here is another:

(2) Let $b \in F$. We say $\lim_{x\to \infty} xa = b$ iff for every positive $\epsilon \in F$, there exists $X \in F$ such that for all $x \in F$, if $x \ge X$, then $|xa-b|<\epsilon$. If there is such an element $b$, then we say $\lim_{x\to\infty} xa$ exists.

(2') We say $\lim_{x\to \infty} xa = \infty$ iff for every $k \in F$, there exists $X \in F$ such that for all $x \in F$, if $x \ge X$, then $xa > k$.

I numbered them (1) and (2) for reference, but there are still more possibilities...

-
With these definitions ... (1) $\lim_{x\to \infty} xa$ does not exist. (1') $\lim_{x\to \infty} xa = \infty$ is false. (2) $\lim_{x\to \infty} xa$ does not exist. (2') $\lim_{x\to \infty} xa = \infty$ is true. –  GEdgar Oct 6 '11 at 15:24
Given that he used the word 'non-standard' and the calculus-oriented questions, I think it would be a reasonable bet that he's asking (or intends to ask) about some version of the hyperreals, rather than some arbitrary ordered field containing $\mathbb{R}$. –  Hurkyl Oct 7 '11 at 7:49

By the transfer principle, any theorem of standard analysis is a theorem of non-standard analysis.

For example, $\lim_{x \to +\infty} b^x = +\infty$ when $b > 1$ is a theorem of standard analysis. Thus, its transfer is a theorem of non-standard analysis. Since $1+a > 1$, we have $\lim_{x \to +\infty} (1+a)^x = +\infty$.

Of course, if $H$ is a positive transfinite number, then $\lim_{x \to H} (1+a)^x$ is not $+\infty$, it is $(1+a)^H$. This value may or may not be transfinite, depending on whether $H$ is large enough (This is closely related to indeterminate forms), But it is definitely smaller than $+\infty$.

The same principle applies to your other examples.

EDIT: with the update (and since Kermit seems to want more comment), we know that $\sum_{n=1}^{+\infty} 1/n^{1+a} = \zeta(1+a)$ and $\sum_{n=1}^{+\infty} a = +\infty$ for all $a > 0$. These statements remain true in the non-standard model.

-
Perhaps the OP should say what he means by those symbols involving $\infty$ in his case. –  GEdgar Oct 6 '11 at 2:46
It would help. I deliberately talked about both the extended (hyper)real $+\infty$ and an ordinary transfinite hyperreal $H$ just in case the OP is mixing the two up. (and to illustrate that even though $H$ is transfinite, it's just an ordinary hyperreal and so you can do things as one would ordinarily do). –  Hurkyl Oct 6 '11 at 9:43
Sorry, does the first sum as of now converge, and to a standard or nonstandard number? –  user1708 Oct 6 '11 at 12:03
What does $\zeta(1+a)$ mean, when $a$ is infinitesimal? –  Gerry Myerson Oct 6 '11 at 12:04
@Gerry: This is what the transfer principle is good for! Since $\zeta$ is a standard function on $\mathbb{C} \setminus \{1\}$, its transfer is an extension to be defined on ${}^\star \mathbb{C} \setminus \{ 1 \}$. Another way of putting it is that the transfer principle ensures everything we do to construct $\zeta$ in the standard case still works in the non-standard case -- e.g. the familiar infinite series formula for real part $> 1$ still works. –  Hurkyl Oct 7 '11 at 2:08