We have spent a whole series of posts adding infinitely many numbers. What happens if we multiply them instead? Infinite products turn out to be close cousins of infinite series, linked by the logarithm, and they give some of the most beautiful formulas for \(\pi\). This post explains when an infinite product converges and how to work with it.
1. Partial products and convergence
The definition copies the one for series, with multiplication in place of addition.
Definition
The infinite product
\[ p_1p_2\cdots p_n\cdots = \prod_{n=1}^\infty p_n \qquad (p_n > 0,\ n = 1, 2, \dots) \tag{1} \]
is convergent if the limit
\[ \lim_{n\to\infty} P_n \]
exists, is finite, and is different from zero, where
\[ P_n = \prod_{k=1}^n p_k = p_1p_2\cdots p_n \]
is the partial product. The limit is called the value of the product.
Why exclude the limit \(0\)? It looks strange at first: if \(P_n \to 0\), the partial products do have a limit. But zero is a "black hole" for multiplication. Once a product gets close to zero, no later factor can say anything meaningful about it, and very different products would all "converge" to \(0\). Excluding it is what makes the link with series work (Section 2): \(P_n \to 0\) corresponds to \( \ln P_n \to -\infty \), a divergent series. So we say a product with \(P_n \to 0\) diverges to zero.
Why \(p_n > 0\)? A single zero factor would make every later partial product zero, and negative factors would make the sign jump around. Requiring positive factors keeps things clean and lets us take logarithms. (More general theories allow finitely many zero or negative factors, but the idea is the same.)
Necessary condition
If the product (1) converges, then \( p_n \to 1 \).
Why: \( p_n = \frac{P_n}{P_{n-1}} \to \frac{P}{P} = 1 \), which uses \(P \ne 0\). This is the twin of the n-th term test for series (\(a_n \to 0\)). For this reason it is customary to write the factors as \( p_n = 1 + a_n \), with \( a_n \to 0 \).
2. The bridge to series: taking logarithms
The logarithm turns products into sums: \( \ln P_n = \ln p_1 + \ln p_2 + \cdots + \ln p_n \). So the partial products of the product are the exponentials of the partial sums of a series.
Theorem
The convergence of the product (1) is equivalent to the convergence of the series
\[ \sum_{n=n_0}^\infty \ln p_n. \tag{2} \]
If the series has sum \(S\), the product has value \( P = e^S \) (when summation starts at \(n_0 = 1\)).
Why: \( P_n = e^{S_n} \), where \(S_n\) are the partial sums of (2). Since \(e^x\) and \(\ln x\) are continuous:
The starting index \(n_0\) doesn't matter: finitely many factors (all positive) only multiply the value by a nonzero constant, just as finitely many terms don't affect the convergence of a series.
This theorem means that everything we know about series applies to products. To test a product, test the series of logarithms.
3. A practical test: products of the form \(\prod(1 + a_n)\)
Logarithms can be awkward to sum. Fortunately, for small \(a\), \( \ln(1 + a) \approx a \) (the differential from an earlier post), and more precisely
\[ \lim_{a\to 0}\frac{\ln(1 + a)}{a} = 1. \]
By the limit comparison test, \( \sum\ln(1 + a_n) \) and \( \sum a_n \) behave the same when the \(a_n\) don't change sign. This gives the most useful test:
Test for products with terms of constant sign
If \( a_n \ge 0 \) for all \(n\) (or \( -1 < a_n \le 0 \) for all \(n\)), then
\[ \prod_{n=1}^\infty(1 + a_n) \ \text{converges} \iff \sum_{n=1}^\infty a_n \ \text{converges}. \]
Example 1 · Two divergent products
\( \prod_{n=1}^\infty\big(1 + \frac1n\big) \): the partial products telescope, \( P_N = \frac21\cdot\frac32\cdots\frac{N+1}{N} = N + 1 \to \infty \). Diverges, as predicted by \( \sum\frac1n = \infty \).
The factors tend to \(1\) in both cases, but not fast enough.
Example 2 · A convergent product
\( \prod_{n=1}^\infty\big(1 + \frac{1}{n^2}\big) \) converges, because \( \sum\frac1{n^2} \) converges. (Its value turns out to be \( \frac{\sinh\pi}{\pi} \approx 3.676 \), which follows from Euler's product for the sine in Section 6.) In general, \( \prod\big(1 + \frac{1}{n^p}\big) \) converges exactly when \(p > 1\), the p-series rule again.
4. Absolute and conditional convergence
Definition
The product (1) is absolutely or conditionally (non-absolutely) convergent according to whether the series (2) converges absolutely or conditionally.
Since \( |\ln(1 + a_n)| \sim |a_n| \), the product \( \prod(1 + a_n) \) converges absolutely exactly when \( \sum|a_n| \) converges. And just as with series (the post on terms of any sign), an absolutely convergent product has a value that does not depend on the order of the factors, while a conditionally convergent one can change its value, or diverge, when the factors are rearranged.
Group the factors in pairs: \( \Big(1 + \frac{1}{2k-1}\Big)\Big(1 - \frac{1}{2k}\Big) = \frac{2k}{2k-1}\cdot\frac{2k-1}{2k} = 1 \). So \(P_{2k} = 1\) and \( P_{2k+1} = 1 + \frac{1}{2k+1} \to 1 \). The product converges to \(1\).
But \( \sum|a_n| = \sum\frac1n = \infty \), so the convergence is only conditional.
With mixed signs, \(\sum a_n\) is not enough
The test of Section 3 needs \(a_n\) of constant sign. Take \( a_n = \frac{(-1)^{n+1}}{\sqrt n} \). Here \( \sum a_n \) converges (Leibniz), yet the product \( \prod(1 + a_n) \) diverges to zero. The reason is the next term of the expansion:
\[ \ln(1 + a) = a - \frac{a^2}{2} + \frac{a^3}{3} - \cdots, \]
and \( \sum a_n^2 = \sum\frac1n = \infty \), so \( \sum\ln(1 + a_n) \) contains a hidden \( -\frac12\sum a_n^2 = -\infty \). In general, if \( \sum a_n \) and \( \sum a_n^2 \) both converge, then \( \prod(1 + a_n) \) converges (as in Example 3, where \(\sum\frac1{n^2} < \infty\)).
5. Computing products: telescoping
As with series, convergence is easier to establish than the value. One technique computes many values exactly: factor each \(p_n\) so that most factors cancel in the partial product.
Example 4 · \( \prod_{n=2}^\infty\big(1 - \frac{1}{n^2}\big) = \frac12 \)
Write \( 1 - \frac1{n^2} = \frac{(n-1)(n+1)}{n\cdot n} = \frac{n-1}{n}\cdot\frac{n+1}{n} \). Then
\[ P_N = \Big(\frac12\cdot\frac23\cdots\frac{N-1}{N}\Big)\Big(\frac32\cdot\frac43\cdots\frac{N+1}{N}\Big) = \frac1N\cdot\frac{N+1}{2} = \frac{N+1}{2N} \longrightarrow \frac12. \]
Notice that each half on its own diverges (to \(0\) and to \(\infty\)), as in Example 1, but together they converge.
Example 5 · A product equal to a fraction
For \(|x| < 1\): \( (1 + x)(1 + x^2)(1 + x^4)\cdots(1 + x^{2^{N}}) = \dfrac{1 - x^{2^{N+1}}}{1 - x} \). (Multiply by \(1 - x\): each step doubles the exponent, \( (1 - x)(1 + x) = 1 - x^2 \), \( (1 - x^2)(1 + x^2) = 1 - x^4 \), and so on.) Letting \(N \to \infty\):
\[ \prod_{n=0}^\infty\big(1 + x^{2^n}\big) = \frac{1}{1 - x}. \]
This is the product version of the fact that every natural number has a unique binary representation.
6. Famous products: Viète, Wallis, Euler
Example 6 · Viète's formula (1593)
The double-angle formula \( \sin x = 2\sin\frac x2\cos\frac x2 \), applied repeatedly, gives
\[ \sin x = 2^N\sin\frac{x}{2^N}\,\cos\frac x2\cos\frac x4\cdots\cos\frac{x}{2^N}. \]
Since \( 2^N\sin\frac{x}{2^N} \to x \) (because \(\frac{\sin t}{t} \to 1\)), for \(x \ne 0\):
\[ \prod_{n=1}^\infty\cos\frac{x}{2^n} = \frac{\sin x}{x}. \]
At \(x = \frac\pi2\), using \( \cos\frac\pi4 = \frac{\sqrt2}{2} \) and the half-angle formula \( \cos\frac\theta2 = \sqrt{\frac{1 + \cos\theta}{2}} \):
\[ \frac2\pi = \frac{\sqrt2}{2}\cdot\frac{\sqrt{2 + \sqrt2}}{2}\cdot\frac{\sqrt{2 + \sqrt{2 + \sqrt2}}}{2}\cdots \]
This was the first infinite expression for \(\pi\) in the history of mathematics.
Euler's product for the sine
Euler discovered that the sine factors over its zeros \( x = \pm\pi, \pm2\pi, \dots \), just like a polynomial factors over its roots:
\[ \sin x = x\prod_{n=1}^\infty\Big(1 - \frac{x^2}{n^2\pi^2}\Big). \]
The product converges absolutely for every \(x\), since \( \sum\frac{x^2}{n^2\pi^2} < \infty \). (The proof uses tools beyond this post, such as Fourier series or complex analysis.)
Example 7 · Wallis' product (1656)
Put \(x = \frac\pi2\) in Euler's product:
\[ 1 = \frac\pi2\prod_{n=1}^\infty\Big(1 - \frac{1}{4n^2}\Big) = \frac\pi2\prod_{n=1}^\infty\frac{(2n-1)(2n+1)}{(2n)^2}. \]
Taking reciprocals:
\[ \frac\pi2 = \prod_{n=1}^\infty\frac{(2n)^2}{(2n-1)(2n+1)} = \frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdots \]
Compare the coefficients of \(x^3\) on both sides of Euler's product (\( -\frac16 \) from the sine series, \( -\frac{1}{\pi^2}\sum\frac{1}{n^2} \) from the product) and you get \( \sum\frac1{n^2} = \frac{\pi^2}{6} \) once more, Euler's original argument.
A glimpse: Euler's product and the primes
For \(s > 1\), expanding each factor as a geometric series and using unique prime factorization gives
\[ \prod_{p\ \text{prime}}\frac{1}{1 - p^{-s}} = \sum_{n=1}^\infty\frac{1}{n^s}. \]
As \(s \to 1^+\) the right side tends to the divergent harmonic series. If there were only finitely many primes, the left side would stay finite. So there are infinitely many primes, proved with an infinite product. This identity is the starting point of analytic number theory.
7. Summary
Series \( \sum a_n \)
Product \( \prod p_n \) (\(p_n = 1 + a_n > 0\))
partial sums \(S_n\)
partial products \(P_n = e^{S_n}\) with \(S_n = \sum\ln p_k\)
Problem 4: Show that \( \prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1} = \dfrac23 \).
\( \frac{n^3 - 1}{n^3 + 1} = \frac{n - 1}{n + 1}\cdot\frac{n^2 + n + 1}{n^2 - n + 1} \). The first factors telescope to \( \frac{1\cdot 2}{N(N+1)} \). For the second, note \( n^2 - n + 1 = (n-1)^2 + (n-1) + 1 \), so they telescope to \( \frac{N^2 + N + 1}{3} \). Hence \( P_N = \frac{2(N^2 + N + 1)}{3N(N+1)} \to \frac23 \).
Problem 5: Is \( \prod_{n=1}^\infty\Big(1 + \frac{(-1)^{n}}{n^2}\Big) \) absolutely convergent?
Yes: \( \sum|a_n| = \sum\frac1{n^2} < \infty \).
Problem 6: Using Viète's product, compute \( \prod_{n=1}^\infty\cos\dfrac{\pi}{3\cdot 2^n} \).
With \( x = \frac\pi3 \): \( \frac{\sin(\pi/3)}{\pi/3} = \frac{\sqrt3/2}{\pi/3} = \frac{3\sqrt3}{2\pi} \).
Problem 7: Why does the product \( \prod_{n=1}^\infty\frac{n}{n+1} \) not converge, even though its partial products have a limit?
\( P_N = \frac{1}{N+1} \to 0 \). By definition, a product converges only if the limit is finite and different from zero. Equivalently, \( \sum\ln\frac{n}{n+1} = -\sum\ln\big(1 + \frac1n\big) = -\infty \) diverges.
Final thoughts
Infinite products look new, but they hide no new theory: take the logarithm, and a product becomes a series. Convergence, absolute convergence, the danger of rearrangement and the comparison with \(\sum a_n\) all come straight from the posts on series. What products add is a different way of seeing functions: Euler's sine product shows a function built from its zeros, and from it flow Wallis' product, Viète's formula and the Basel sum, some of the most elegant results in all of analysis.