Infinite Products - Revisited

Here we look again at infinite products because previously we didn't focus on infinite products of complex numbers, and presented the convergence criteria without explaining them.

The video for this topic is at [youtube], and slides are here [pdf].

Initial Intuition

Let's first develop an intuition for infinite products through some examples.

$$2\times 3\times 4\times 5\times \dots$$

It is easy to see the above product diverges. Each factor increases the size of the product. 

$$2\times 0\times 4\times 5\times \dots$$

It is a fundamental idea that multiplying by zero causes a product to be zero. The product is zero because one of the factors is zero.

$$\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{1}{5}\times \dots$$

This product is more interesting. Each factor is a fraction that reduces the size of the product. As the number of these reducing factors grows, the product gets ever closer to zero. We can make the leap to say the value of the infinite product is zero.

We have found two different ways for the product to be zero. We'll need to keep both in mind as we work with infinite products. 

Definition

Similar to infinite series, we say an infinite product converges if the limit of the partial products is a finite value.

$$\underset{N\to \mathrm{\infty }}{lim}\prod _{n=1}^N{a}_n=P$$

We'll see why it is conventional to insist the finite value is non-zero.

Example 1

Does the infinite product $\prod _{n=1}^{\mathrm{\infty }}(1+1/n)$ converge? 

Each factor $(1+1/n)$ is larger than one, so we expect the product to keep growing. Let's be more rigorous and see how the partial products actually grow.

$$\begin{array}{rl}\prod _{n=1}^N(1+\frac{1}{n})& =\prod _{n=1}^N\left(\frac{n+1}{n}\right)\\ & \\ & =\frac{\cancel{2}}{1}\times \frac{\cancel{3}}{\cancel{2}}\times \frac{\cancel{4}}{\cancel{3}}\times \dots \times \frac{N+1}{\cancel{N}}\\ & \\ & =N+1\end{array}$$

As $N\to \mathrm{\infty }$, the product diverges.

Example 2

Lets now look at the the similar infinite product $\prod _{n=2}^{\mathrm{\infty }}(1-1/n)$. Notice n starts at 2 to ensure the first factor is not zero. 

Each factor $(1-1/n)$ is smaller than one, so we expect the product to keep shrinking. Let's see if the partial products do indeed get smaller. 

$$\begin{array}{rl}\prod _{n=2}^N(1-\frac{1}{n})& =\prod _{n=2}^N\left(\frac{n-1}{n}\right)\\ & \\ & =\frac{1}{\cancel{2}}\times \frac{\cancel{2}}{\cancel{3}}\times \frac{\cancel{3}}{\cancel{4}}\times \dots \times \frac{\cancel{N+1}}{N}\\ & \\ & =\frac{1}{N}\end{array}$$

As $N\to \mathrm{\infty }$, the product tends to zero. Remember that for convergence we insist the limit is non-zero. For this reason we say the product diverges to zero.

Convergence And $a_n$

We know that for an infinite series $\sum a_n$ to converge, the terms $a_n$ must $\to 0$. For an infinite product $\prod a_n$ to converge, the terms $a_n\to 1$. 

If each term $a_n$ was larger than 1, the product would get ever larger. If each term $a_n$ was smaller than 1, the product would get ever smaller towards zero. Negative $a_n$ cause partial product to oscillate, meaning convergence only happens if $a_n\to 1$.

Removing Zero-Valued Factors

A single zero-valued factor collapses an entire product to zero. If an infinite product has a finite number of zero-valued factors, they can be removed to leave a potentially interesting different product. 

For example, the following product is zero because the first factor is zero. 

$$\prod _{n=1}(1-\frac{1}{n^2})=0$$

Removing the first factor leaves a much more interesting product.

$$\prod _{n=2}(1-\frac{1}{n^2})=\frac{1}{2}$$

Convergence Criteria 1

Since the terms in a convergent infinite product tend to 1, it is useful to write the factors as $(1+a_n)$. 

$$P=\prod (1+a_n)$$

We can turn a product into a sum by taking the logarithm.

$$\ln (P)=\ln \prod (1+a_n)=\sum \ln (1+a_n)$$

Using $1+x\le e^x$ we arrive at a nice inequality.

$$\ln (P)\le \sum a_n$$

This tells us that if the sum is bounded, the product is bounded too. If the terms $a_n$ are always positive, then the sum can only grow monotonically (without oscillation), so the boundedness is convergence. This is a useful result but we can strengthen it.

Expanding out the product $\prod (1+a_n)$ gives us a sum which includes the terms 1, all the individual $a_n$, and also the combinations of different $a_n$ multiplied together. This gives us an inequality for $\sum a_n$ in the other direction.

$$1+\sum a_n\le \prod (1+a_n)=P$$

This tell us that if the product converges, so does the sum. The two results together give us our first convergence criterion.

$$\sum a_n\text{ converges }\iff \prod (1+a_n)\text{ converges, for }{a}_n>0$$

This allows us to say $\prod (1+1/n)$ diverges because $\sum 1/n$ diverges, and that $\prod (1+1/n^2)$ converges because $\sum 1/n^2$ converges.

Convergence Criterion 2

A very similar argument that uses $1-x\le e^{-x}$ leads to a criterion for products of the form $\prod (1-a_n)$.

$$\sum a_n\text{ converges }\iff \prod (1-a_n)\text{ converges, for }0<a_n<1$$

So $\prod _2^{\mathrm{\infty }}(1-1/n)$ diverges, because $\sum 1/n$ diverges.

Divergence To Zero

The logarithmic view of infinite products is useful because it turns a product into a sum, but it has an interesting side effect. 

If the partial products tend to zero, then the logarithm diverges towards $-\mathrm{\infty }$. This is why we say a product diverges to zero. 

Convergence Criteria 3

The previous convergence criteria are for real values of $a_n$. It would be useful to have a criterion for complex $a_n$. To do that we need an intermediate result about $|a_n|$.

For complex $a_n$, we have $|a_n|>0$ for all $a_n\ne 0$. This gives is an intermediate result. 

$$\sum |a_n|\text{ converges }\iff \prod (1+|a_n|)\text{ converges}$$

We are interested in products $\prod (1+a_n)$ with complex $a_n$, not just $\prod (1+|a_n|)$. Let's start with two partial products with complex $a_n$.

$$p_N=\prod ^N(1+a_n)$$

$$q_N=\prod ^N(1+|a_n|)$$

We need to assert $a_n\ne -1$ to ensure no zero-valued factors $(1+a_n)$. 

For $N>M\ge 1$, we can compare $|p_N-p_M|$ with $|q_N-q_M|$ with a little algebra.

$$\begin{array}{rl}|p_N-p_M|& =|p_M|\cdot |\frac{p_N}{p_M}-1|\\ & \\ & =|p_M|\cdot |\prod _{M+1}^N(1+a_n)-1|\\ & \\ & \le |q_M|\cdot |\prod _{M+1}^N(1+|a_n|)-1|\\ & \\ & =|q_M|\cdot |\frac{q_N}{q_M}-1|\\ & \\ |p_N-p_M|& \le |q_N-q_M|\end{array}$$

If $|q_N-q_M|<ϵ$, where $ϵ$ is as small as we want, then $|p_N-p_M|<ϵ$ too. This the Cauchy criterion for convergence, and it tells us that if $q_N$ converges, so does $p_N$.

So $\sum |a_n|$ converges means $\prod (1+|a_n|)$ converges, which we can now say means $\prod (1+a_n)$ also converges. We finally have our third convergence criterion.

$$\sum |a_n|\text{ converges }⟹\prod (1+a_n)\text{ converges, for }{a}_n\ne -1$$

Notice this criterion is one way. We can't say the sum converges if the product converges. So we've have a new constraint.

Why Convergence Is Non-Zero

Let's see why convergence according to these criteria mean the products converge to a non-zero value. We've already seen how $a_n\to 0$, which means that $|a_n|<1/2$ except for a finite number of terms. 

We use the useful inequality $1+x\le e^x$ again.

$$1\le \prod (1+|a_n|)<e^{\sum |a_n|}$$

This tells us that if the sum $\sum |a_n|$ converges, then the product $\prod (1+|a_n|)$ converges and is non-zero. 

We can use another inequality $1-x\ge e^{-2x}$ for $0\le x\le 1/2$, and that $e^y>0$ for all real $y$.

$$0<e^{-2\sum |a_n|}\le \prod (1-|a_n|)\le 1$$

This tells us that if the $sum\sum |a_n|$ converges, then the product $\prod (1-|a_n|)$ converges and is non-zero. 

Now we use another inequality $1-|a_n|\le |1\pm a_n|\le 1+|a_n|$ to relate $\prod (1-|a_n|)$ to $\prod (1-a_n)$.

$$\prod (1-|a_n|)\le |\prod (1-a_n)|\le \prod (1+|a_n|)$$

Assuming $\sum |a_n|$ converges, we can finally say $|\prod (1-a_n)|$ is non-zero, because its value is between two known non-zero values.

Riemann Zeta Function $\zeta (s)\ne 0$ for $\sigma >1$ 

The Riemann Zeta function can be written as an infinite product over primes. Here $s=\sigma +it$, and $\sigma >1$.

$$\zeta (s)=\sum \frac{1}{n^s}=\prod {(1-\frac{1}{p^s})}^{-1}$$

It is natural to ask if $\zeta (s)$ has any zeros in the domain $\sigma >1$.

None of the factors $(1-1/p^s{)}^{-1}$ is zero. That would require $p^s$ to be zero, and that isn't possible. 

$$\left|p^s\right|=\left|e^{s\ln (p)}\right|=e^{\sigma \ln (p)}>0$$

We also need to check the infinite product doesn't diverge to zero. For the moment let's consider $1/\zeta (s)=\prod (1-1/p^s)$. Using the third convergence criterion, we check whether $\sum |1/p^s|$converges.

$$\sum \left|\frac{1}{p^s}\right|=\sum \frac{1}{p^{\sigma }}\le \sum \frac{1}{n^{\sigma }}$$

The reason $\sum 1/p^{\sigma }\le \sum 1/n^{\sigma }$ is because there are fewer primes $p$ than integers $n$. Because $\sum 1/n^{\sigma }$ converges for $\sigma >1$, so does $\sum |1/p^s|$. This means $1/\zeta (s)$ converges to a non-zero value, and therefore so does $\zeta (s)$.

We can now say the Riemann Zeta function has no zeros in the domain $\sigma >1$.