Convergence Of Infinite Products

Infinite Series (Sums)

At school we learned about infinite series.

$$  \sum_{n=1}^\infty a_n = a_1+ a_2 + a_3 + \ldots  $$

If the partial sum $\sum_{n=1}^{x} a_n$ tends to a limit as $x \rightarrow \infty$, we can say the series converges to a finite sum. There are many tests that can be used to see if a series converges or not, such as the popular ratio test:

$$ \lim_{n \rightarrow \infty} | \frac{a_{n+1}}{a_n} | <1 $$

If we think about the series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$, the ratio test gives us $|\frac{ 2^{n-1} }{ 2^{n} }| = \frac{1}{2}$, confirming the series is absolutely convergent.

Infinite Products

Infinite products don't seem to be discussed as much as series.

$$ \prod_{n=1}^\infty a_n = a_1 \times a_2 \times a_3 \times \ldots  $$

When do such infinite products converge? Let's think about this before jumping to the answer.

Thoughts

It seems intuitive that if the elements $a_n$ are getting larger, the product grows and diverges. The following example gets ever larger and doesn't converge. 

$$  2 \times 3 \times 4 \times 5 \times \ldots $$

Let's look at another example.

$$ 0.5 \times 0.55 \times 0.555 \times 0.5555 \times \ldots $$

Here the elements are getting larger, but are individually less than 1, so it isn't immediately obvious whether the product converges or not.

What if the product contains a zero? That sounds easy. A single zero would immediately make the overall product zero. 

But what about this product? 

$$ -2 \times -1 \times 0 \times 1 \times 2 \times \ldots $$

With the zero, we'd suggest the product is zero overall. But without that zero, the product gets larger and diverges. Is the single zero somehow 'powerful' enough to cancel the infinity?

Definition

A consensus has emerged around infinite products which may, or may not, seem natural.

If the partial product $\prod_{n=1}^{x} a_n$ converges to a non-zero limit $P$ as $x \rightarrow \infty$, we say the infinite product converges to $P$.

This makes sense but the fact that the limit needs to be non-zero sets infinite products apart from infinite series.

If a finite number of zeros $a_n$ can be removed from a product, and the remaining product converges, we say the infinite product converges to 0.

This is rather specific, and means that we can't say a product with infinitely many zero elements converges. For example, we can't say the following converges:

$$  1 \times 0 \times 1 \times 0 \times 1 \times 0 \times 1 \times \ldots $$

Let's look at some examples to bring these ideas to life.

Example 1

Does the following infinite product converge or diverge?

$$ \prod_{n=1}^{\infty}(1+\frac{1}{n}) $$

If we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=1}^{N}(1+\frac{1}{n}) &=  \prod_{n=1}^{N}(\frac{n+1}{n}) \\ \\ &= \frac{\cancel{2}}{1} \times \frac{\cancel{3}}{\cancel{2}} \times \frac{4}{\cancel{3}} \times  \ldots \times \frac{N+1}{N} \\ \\ &= N + 1 \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to $\infty$, so the infinite product diverges. This isn't surprising when we look at each element and that they are all above 1.

Example 2

Let's now look at an example where each element is less than 1. Notice $n$ starts at 2, not 1. 

$$ \prod_{n=2}^{\infty}(1-\frac{1}{n}) $$

Again, if we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=2}^{N}(1-\frac{1}{n}) &=  \prod_{n=2}^{N}(\frac{n-1}{n}) \\ \\ &= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{4} \times  \ldots \times \frac{N-1}{N} \\ \\ &= \frac{1}{N} \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to 0. Because this limit is not a non-zero limit, as required by the definitions above, we say the infinite product diverges. 

Turning Products Into Sums

We can convert a product into a sum using logarithms.

$$ \begin{align} \ln (\prod_{n=1}^\infty a_n) &= \ln( a_1 \times a_2 \times a_3 \times \ldots)  \\ \\ &= \sum_{n=1}^\infty \ln(a_n) \end{align}$$

We can say the infinite product converges if the sum $\sum_{n=1}^\infty \ln(a_n)$ converges. 

If that product tends to zero, this sum tends to $-\infty$, which is why we say the infinite product diverges to zero, not converges to zero.

A Different Test For Convergence

If we can rewrite the product elements as $(1 + a_n)$ where $a_n >0$ we can use a simpler convergence test.

$$ \begin{align} \prod_{n=1}^\infty (1 + a_n) &\leq  \prod_{n=1}^\infty e^{ a_n} \\ \\ &=   e^{\sum_{n=1}^\infty a_n} \end{align} $$

The inequality comes from $e^x=1 + x + O(x^2)$. We can make the following conclusion.

If $\sum_{n=1}^\infty a_n$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n >0$.

This test allows us to easily say the following infinite series converges:

$$ \prod_{n=1}^\infty (1 + \frac{1}{n^2}) $$

We can say this because we know $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.

Infinite Products Over Complex Numbers

There is an equivalent form of the previous test for complex $z_n$.

If $\sum_{n=1}^\infty |a_n|$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n \neq -1$.