Is it true that infinite product of infinitesimals may not be an infinitesimal? Is it true that infinite product of infinitesimals may not be an infinitesimal?
Here's my attempt.
Using ultraproduct construction,
 $$\epsilon_1=[\langle 1, \frac{1}{2}, \frac{1}{3},\frac{1}{4},\ldots\rangle]$$
$$\epsilon_2=[\langle 1, 2, \frac{1}{3},\frac{1}{4},\ldots\rangle]$$
$$\epsilon_3=[\langle 1, 1, 9,\frac{1}{4},\ldots\rangle]$$
$$\ldots$$
In general, $\epsilon_n=[\langle a_k\rangle_{k \in \Bbb N}]$ such that : $$a_k =
\begin{cases}
1,  & \text{if $k < n-1$ } \\
n^{n-1}, & \text{if $k=n-1$ }  \\
\frac {1}{n+1}, & \text{if $k > n-1$ }  \\
\end{cases}$$
So $\prod_{i \in {\Bbb Z}_{+}}{\epsilon_i} = 1$. I think it's true, but it's counterintuitive. Is there something wrong? Or is there something subtle that prevent it hehaving like the limit of a standard Cauchy sequence? Is there an standard example?
Added: As pointed out in  Brian M. Scott's comment, the problem is that "componentwise products cannot be right", different representatives of a hyperreal will give different answers. So is there a correct definition of infinite product of hyperreals?
 A: A sequence $\langle \epsilon_n : n\in\mathbb{N}\rangle$ can be extended to an internal sequence $\langle \epsilon_n : n\in{}^\ast\mathbb{N}\rangle$ of infinitesimals. This is a nontrivial result using saturation (it should be in Goldblatt though I can't provide a page number right now). One can interpret an infinite product as a hyperfinite product of the extended sequence up to hyperfinite rank $H\in{}^\ast\mathbb{N}\setminus\mathbb{N}$.  Then, since all the factors are infinitesimal, by the transfer principle the product will be less than the first factor, which is infinitesimal, and therefore the product itself is infinitesimal.
A: This one is easy to see if we take a nonstandard model of real analysis, rather than just a nonstandard model of the reals.
In standard analysis, for every natural number $n$, there is a set $S_n$ of all $n$-long real sequences. This can be viewed as a function whose domain is $\mathbb{N}$ and whose codomain is an irritating but straightforward to describe set.
Furthermore, we have the product operator from sequences to the reals: given any such sequence $\{ s_k \}$, we have a real number $\prod s$
So when we transfer to the nonstandard model, this extends. For each hypernatural number $n$ -- even transfinite ones -- we have an internal, nonstandard set $S_n$ of all (internal) $n$-long hyperreal sequences. And we have a product operator, so that for any such sequence $\{ s_k \}$, we have a hyperreal number $\prod s$.
For the standard product operator, we can prove that $|\prod s| < s_0$ if $|s_k| < 1$ for all $k$. And so this remains true for the transfer.
In particular, we see that any hyperfinite product -- even a transfinite product -- of an internal sequence of infinitesimals is defined, and must be infinitesimal.

None of this will work for an external sequence of hyperreal numbers. In particular, we have no reason to expect there to be any reasonable notion of an infinite product
$$ \prod_{k=1}^{+\infty} x_k $$
when $\{ x_k \}$ is a sequence of hyperreals indexed by the ordinary natural numbers.
A: I do not think your example is an infinite product of infinitesimals.
For example,
$$
\begin{align}
S=&\lim_{n\to\infty}\sum_{k=1}^n\frac k{n^2}\\
=&\lim_{n\to\infty}\left(\frac1{n^2}+\frac2{n^2}+\cdots+\frac{n-1}{n^2}+\frac n{n^2}\right)\\
=&\lim_{n\to\infty}\frac{\frac12n(n+1)}{n^2}=\frac12
\end{align}
$$
is a widely accepted infinite sum of infinitesimals. We can see that any term in the sum is an infinitesimal, including $\frac n{n^2}$.
Set $a_k(n)=\frac{k}{n^2}$, the example satisfies
$$
\lim_{n=\infty}a_1(n)=\lim_{n=\infty}a_2(n)=\cdots=\lim_{n=\infty}a_{n-1}(n)=\lim_{n=\infty}a_n(n)=0
$$
and
$$
S=\lim_{n\to\infty}\sum_{k=1}^n a_k(n).
$$
Similarly, I call such
$$
P=\lim_{n\to\infty}\prod_{k=1}^n a_k(n)
$$
as an infinite product of infinitesimals.
Obviously, your example does not satisfy this requirement. $n^{n-1}$ is always a term in the product, but it is not an infinitesimal, which is counterintuitive as you said.
However, I cannot provide an example that $P\ne0$, and I cannot either prove that $P$ must be $0$. I guess $P$ must be $0$.
