Convergence of Infinite Products « The Everything Seminar

Convergence of Infinite Products

There is a simple convergence test for infinite products that I think deserves to be better known.

Theorem. Let $a_n$ be a sequence of positive numbers. Then the infinite product

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

converges if and only if the series

$\displaystyle\sum_{n=1}^{\infty} a_n$

converges.

Proof: Taking the logarithm of the product gives the series

$\displaystyle\sum_{n=1}^{\infty} \ln(1 + a_n)$ ,

whose convergence is equivalent to the convergence of the product. But observe that

$\displaystyle\lim_{x\rightarrow 0} \frac{\ln(1+x)}{x} = 1$ .

If we assume that $a_n \rightarrow 0$ , this gives us that

$\displaystyle\lim_{n\rightarrow \infty} \frac{\ln(1+a_n)}{a_n} = 1$ ,

and the theorem follows by the limit comparison test. Q.E.D.

Using this theorem, everything you know about infinite series translates directly to the world of infinite products. For example, the product

$\displaystyle \prod_{n=1}^{\infty} \left( 1 + \frac{1}{n^p} \right)$

converges if and only if $p > 1$ .

Before I learned this theorem, I had imagined that there must be an entire theory of convergence for infinite products, as complex and interesting as the theory of series from calculus, but completely unknown to me. Instead, it turns out that no one ever talks about the convergence of infinite products because there is basically nothing new to say!

The Harmonic Series
Another reason I like this theorem is that it gives a nice proof that the harmonic series diverges. According to the theorem, the behavior of the harmonic series is the same as the behavior of the following product:

$\displaystyle\left(1 + 1\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right)\cdots$

But this is just

$\displaystyle\frac{2}{1}\,\times\,\frac{3}{2}\,\times\,\frac{4}{3}\,\times\,\frac{5}{4}\,\times\,\cdots$

This clearly diverges, for the partial products are the sequence of positive integers.

Problems
Finally, here’s a fun little pair of exercises:

1. Find a sequence $a_n$ of real numbers such that $\sum a_n$ converges but $\prod (1 + a_n)$ diverges.

2. Find a sequence $a_n$ of real numbers such that $\sum a_n$ diverges but $\prod (1 + a_n)$ converges (and is greater than zero).

This entry was posted on January 26, 2008 at 7:18 am and is filed under Basic Grad Student, High School, Jim, Undergraduate. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

9 Responses to “Convergence of Infinite Products”

Isabel Lugo Says:
January 26, 2008 at 10:21 am
People don’t know this? I would have thought it was common knowledge.

A similar related result is used often in probability. The result is that for real numbers with $0 \le a_n \le 1$ , the infinite product $\prod_{n=1}^\infty (1-a_n)$ converges to a nonzero real number if and only if the sum $\sum a_n$ converges. The proof is essentially the same as the one given above.

One place in which products like this come up fairly often is in situations in probability in which one has a countably infinite number of independent events that happen with small probability (say $a_i$, for i = 1, 2, 3, … and you want to know the probability that none of them happens. This is just $\prod_{i=1}^\infty (1-a_i)$.

(The form I gave comes up much more often than the form you gave in probabilistic contexts, for the obvious reason that probabilities are numbers between zero and one.)
John Armstrong Says:
January 26, 2008 at 11:37 am
Indeed, I thought this was standard. It was certainly one of the first things that Lang screamed at us in intermediate complex analysis.

Although I also have to agree with Jim on one point. I used to wonder whether there was a whole theory of infinite products as well.
Terence Tao Says:
January 26, 2008 at 1:30 pm
The existence of the logarithm function does make the theory of infinite products of scalars essentially equivalent to the theory of infinite series, but the subject becomes significantly richer when one works with infinite products of matrices or operators. Indeed, this is essentially the theory of discrete linear systems $x_{n+1} = A_n x_n$ , where the $x_n$ are vectors and $A_n$ are matrices, which can be viewed as a discretised model for linear non-autonomous ODE. Even the $2 \times 2$ case is of significant interest (in the theory of Schrodinger and Dirac equations, or more generally in understanding flows on $SL_2({\Bbb R})$ ). One useful principle in this subject goes by the colourful name of the “avalanche principle”, and roughly speaking asserts that if $A_1, A_2, \ldots \in SL_2({\Bbb R})$ is such that the operator norm of $A_i A_{i+1}$ is close to the product of the operator norms of $A_i$ and $A_{i+1}$ , then the operator norm of $A_1 \ldots A_n$ is close to the product of the operator norms of $A_1,\ldots,A_n$ , thus in this case at least one can reduce the matrix product problem to a scalar product problem. But there are certainly other more oscillatory scenarios in which matrix products behave very differently from scalar products.

Incidentally, your proof of divergence of the harmonic function also gives the right asymptotic $1+\frac{1}{2}+\ldots+\frac{1}{n} = \log n + O(1)$ for the partial sums. By working slightly harder, it also gives a proof of existence of Euler’s constant $\gamma$ , though it does not give a particularly useful formula for what that constant is.

p.s. I believe Isabel wanted to write $a_n < 1$ rather than $a_n \leq 1$ .
Jim Belk Says:
January 26, 2008 at 4:55 pm
People don’t know this? I would have thought it was common knowledge.

I managed to not know this until my third or fourth year of graduate school. Of course, I was mostly studying algebra and topology — it’s presumably common knowledge for those who work in probability or other fields where this comes up.

I guess my complaint is: Why isn’t this in calculus books? Or at least undergraduate analysis books? It’s such a simple result, and it conveys so much information about infinite products, that it ought to be better advertised.
Greg Muller Says:
January 26, 2008 at 5:56 pm
I have very fond memories of this fact, since it was posed as an extra-text-ual homework problem in an undergraduate analysis class. I solved it while riding one of the rutgers buses around and around in a circle. It was only a month or so after I’d started taking math classes because engineering was pissing me off so much, and it the fun of solving it helped convince me that math was for me.

Still, it was assigned as one of those classroom asides of the “everyone should know this but its not in your textbook” variety, so the teacher agreed that its not as canonical a fact as it should be.
Isabel Lugo Says:
January 26, 2008 at 8:08 pm
Indeed, I should have excluded $a_n = 1$.
yaroslavvb Says:
January 27, 2008 at 3:05 am
An interesting related case is convergence of infinite products of non-negative matrices in rank. IE, under which conditions is the rank of infinite product 1? This is important in signal processing and graphical models — since algorithms like filtering for HMMs, sum-product are essentially just a sequence of non-negative matrix multiplications, convergence to rank 1 means initial conditions are forgotten and the algorithms are stable.

One result is — product of matrices $A_1,A_2\ldots$ converges to rank 1 (is ergodic) if
$\sum_{k=1}^\infty \sqrt{\phi(A_k)}=\infty$
Where $\phi(A)$ is the minimum of cross-ratio $\frac{b_{11}b_{22}}{b_{12}b_{21}}$ taken over all 2×2 submatrices of A.
Omar Antolín Camarena Says:
January 27, 2008 at 2:23 pm
Seriously? This isn’t common knowledge? That is so weird.

I was taught this result as an undergraduate at least three times: once in Calculus, once in Analysis and once in Complex Analysis.
Primitive root mod p Says:
February 27, 2008 at 6:23 pm
[...] the post Convergence of Infinite, we see a simple but strong convergence [...]

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Convergence of Infinite Products

9 Responses to “Convergence of Infinite Products”

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