# What is the product of infinitely many infinitesimals?

In general, taking the sequence for example, if $$\lim\limits_{n \to \infty}a(n)=0$$, we call the sequence $$a(n)$$ is an infinitesimal.

It's well known that, the product of a finite number of infinitesimals is still an infinitesimal, which can be proven by induction. Suppose that $$\lim\limits_{n \to \infty}a_1(n)=\lim\limits_{n \to \infty}a_2(n)=0$$. Then according to the rule of the limits product, $$\lim\limits_{n \to \infty}[a_1(n)a_2(n)]=0$$, which shows that the product of two infinitesimals is an infinitesimal. Thus, by induction, we can generalize the conclusion to the case when a finite number of infinitesimals multiply.

But what about the product of infinitely many infinitesimals? How to define such a product?

• You can cook up an example where the limit of the infinite product is anything you want - have the decay of the first n be cancelled out by the next n. For example, set the first n to to 1/n, and the next n to n , and then the rest to one. Each one converges to zero (after maybe getting very large), and the product of all of them is 1. – Lorenzo Najt Dec 19 '18 at 6:05
• can you elaborate your example in details ? – mengdie1982 Dec 19 '18 at 6:28
• Define $a_i(n)$ ($n$ is the time variable) by setting it to be $1$ unless $i <= 2n$, at which time it is $1/n$ if $i <= n$ and otherwise it is set to $n$. Since eventually $n >= i$, each sequence $a_i$ looks eventually like the sequence $1/n$ so it converges to zero. The product at each time is 1 (if not adjust definition slightly), so the limit of the product is 1. At each time the infinite product is a finite product, since all but finitely many terms are 1. – Lorenzo Najt Dec 19 '18 at 17:15

Notice that there are $$k$$ terms greater than $$\dfrac{1}{2}$$ in $$\{x_n^{(k)}\}_{n=1}^{\infty}$$. Then let $$k \to \infty$$, there are infinitely many terms greater than $$\dfrac{1}{2}$$. As result, $$\prod\limits_{k=1}^{\infty} x_n^{(k)}$$ is not the product of infinitely many infinitesimals.