Convergence of Infinite Products

by Jim Belk

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.

52 Responses to “Convergence of Infinite Products”

Isabel Lugo Says:
January 26, 2008 at 10:21 am | Reply
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 | Reply
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 | Reply
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 $a_n \leq 1$ .
- Michael Hardy Says:
  November 30, 2017 at 12:39 pm | Reply
  Here you’re writing about infinite products of matrices, but how about something that bears the same relation to matrix multiplication that Riemann sums bear to addition?
Jim Belk Says:
January 26, 2008 at 4:55 pm | Reply
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 | Reply
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 | Reply
Indeed, I should have excluded $a_n = 1$.
yaroslavvb Says:
January 27, 2008 at 3:05 am | Reply
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 | Reply
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.
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February 27, 2008 at 6:23 pm | Reply
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Jordan Bell Says:
July 27, 2008 at 12:18 am | Reply
Sigh. I get depressed that so many mathematicians consider most things they know common knowledge. I use intricate arguments involving infinite products all the time, but I remember how mysterious they seemed when I first saw them. I saw them as an undergraduate doing reading on my own, and I wasn’t “taught” them until a graduate complex analysis course.
ken Says:
October 13, 2008 at 1:18 am | Reply
Do you have a link for this paper? Does this paper include the case when the infinite product converges to rank 1 and each row of the final matrix is infinite?

By the way, is it possible to extend the case to an n\times n case. Also, do you have some papers talking about the application of the infinite product in HMM?

Thanks a lot.

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.
Amin Says:
February 3, 2009 at 12:35 am | Reply
Hi
I am wondering what is the value of $\prod_{i=1}^\infty (1-\dfrac{1}{2^{i}})$.
Can any one help me on this?

Thanks
Michael Lugo Says:
February 12, 2009 at 11:17 pm | Reply
Amin,

that product is approximately 0.2887880951; it doesn’t seem to be any well-known constant.

One way to see this is just to multiply out the first thirty or so terms. Another way is to note that

-\log (1 – 1/2^i) = \sum_{n=1}^\infty 1/(n 2^{ni})

and so the negative log of your product is given by

\sum_{i=1}^\infty sum_{n=1}^\infty 1/(n 2^{ni})

Since all terms here are positive the two sums can be interchanged; thus we have

sum_{n=1}^\infty 1/n \sum_{i=1}^\infty 1/2^{ni}

and summing the geometric series, this is

sum_{n=1}^\infty 1/(n(2^n-1))

Of course this then has to be worked out; it’s about c = 1.242062095 and your sum is e^{-c}. But it’s nice to have the transformation into an infinite sum because a lot of people are more comfortable with infinite sums than with infinite products.
- Alex Glazkov Says:
  November 20, 2016 at 10:54 pm | Reply
  What do you mean this is not a well-known constant?
  Have you seen this – https://oeis.org/A048651 ?
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Scott Carnahan Says:
June 26, 2009 at 10:20 pm | Reply
I don’t have any math to contribute (aside from noting that I find this convergence criterion useful in my work), but I’d like to say that I agree with Jordan Bell’s comment. Assertions of obviousness or common knowledge are among the most annoying behaviors I encounter in the math community. Of course, it isn’t limited to math, but that’s where I see it a lot.
Richard Says:
September 8, 2009 at 7:37 am | Reply
Thanks for all this. It is closely related to my
Of finding the probability that a simulated anneal
fails to find a nearby step function.
Aaron F. Says:
December 27, 2009 at 3:06 am | Reply
I’d like to throw myself in with the few benighted souls who didn’t know that this was common knowledge. 🙂 Of course, my undergraduate education in analysis was woefully incomplete…
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January 9, 2010 at 7:55 pm | Reply
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ganeshsree Says:
April 7, 2010 at 1:49 am | Reply
hi…i’m an undergraduate student taking complex analysis this semester…can someone help me with these questions (my professor gave it to us as exercises and i have no idea how to solve it):

A rational function whose only pole is at infinity is a polynomial. Why?

If f(z) is uniformly continuous in a domain D then f(z) is continuous in D. Can we prove this statement? If yes, please prove it? Thank you so so much!
CogitoErgoCogitoSum Says:
October 24, 2010 at 8:46 pm | Reply
Problem!

Whats the deal with your two questions at the end? Find a series that converges/diverges even though the associated product diverges/converges.

I dont know anything about finding convergence of infinite products, but I do know that if there are solutions to your questions then you are in contradiction.

The very first theorem you posted says that sums and associated products both converge and both diverge… together. You used the “if and only if” operator to make both equivalent.
CogitoErgoEst Says:
February 27, 2011 at 12:54 pm | Reply
@CogitoErgoCogitoSum

Firstly, thanks for being the first to comment on the two questions at the end.

Secondly, thanks for the inspiration that helped me think up a pseudonym for this post.

Thirdly, you’re not necessarily correct in saying that Jim is contradicting himself. His hypothesis for the stated theorem is, “Let a_n be a sequence of positive numbers…”, but the problem just says ” … a sequence of (real) numbers … “, which isn’t the same thing.

That would suggest sequences of mixed sign as solutions to the two problems, though I have yet to find a solution to either of them.
- CogitoErgoCogitoSum Says:
  November 14, 2011 at 6:33 am | Reply
  @CogitoErgoEst. I must contest, actually. You are right that the questions at the end ask for any real or complex number sequence as solutions. However, the PROOF of the theorem rests on the premise that the sequences are positive and real. Therefore we would be incorrect to use the theorem for any other type of sequence.
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