Let \(p(n)\) be the number of partitions of \(n\) into unequal parts.
Show that \[ f(z)=\prod_{n=1}^{\infty}(1+z^n) \] defines an analytic function on the unit disk \(\Delta\).
Show that \[ f(z)=\sum_{n=0}^{\infty}p(n)z^n. \]
Show that \(f\) cannot be analytically continued beyond the unit disk: if \(g\) is analytic on a connected domain \(U\supset\Delta\) and \(g=f\) on \(\Delta\), then \(U=\Delta\).
Solution
We use the convention \(p(0)=1\), corresponding to the empty partition.
1. The product\[ f(z)=\prod_{n=1}^{\infty}(1+z^n) \]converges locally uniformly on \(\Delta\) and defines an analytic functionthere.
Fix \(0<\rho<1\). On \(|z|\le\rho\) we have \[ \sum_{n=1}^{\infty}|z|^n \le \sum_{n=1}^{\infty}\rho^n<\infty. \] Choose \(N_0\) so that \(\rho^n\le1/2\) for \(n\ge N_0\). For \(|w|\le1/2\), the branch of \(\log(1+w)\) with value \(0\) at \(w=0\) satisfies \[ |\log(1+w)|\le 2|w|. \] Hence \[ \sum_{n=N_0}^{\infty}\log(1+z^n) \] converges uniformly on \(|z|\le\rho\) by the Weierstrass \(M\)-test. Therefore \[ \prod_{n=N_0}^{\infty}(1+z^n) =\exp\!\left(\sum_{n=N_0}^{\infty}\log(1+z^n)\right) \] is analytic there. Multiplying by the finite product over \(n<N_0\) proves that \(f\) is analytic on every compact subdisk, hence on \(\Delta\).
2. For every \(z\in\Delta\),\[ f(z)=\sum_{m=0}^{\infty}p(m)z^m. \]
For a finite product, \[ \prod_{n=1}^{N}(1+z^n) =\sum_{A\subseteq\{1,\ldots,N\}}z^{\sum_{a\in A}a}. \] Choosing the term \(z^n\) from a factor means choosing the part \(n\), and each factor can be chosen at most once. Thus the coefficient of \(z^m\) is the number of partitions of \(m\) into unequal parts whose parts are at most \(N\).
For fixed \(m\), this coefficient stabilizes to \(p(m)\) once \(N\ge m\). Moreover, for \(|z|<1\) the sum over all finite subsets is absolutely convergent, since \[ \sum_{A\subset\mathbb N\,\text{finite}}|z|^{\sum_{a\in A}a} =\prod_{n=1}^{\infty}(1+|z|^n)<\infty \] by 1 applied to the positive real number \(|z|\). We may therefore group terms according to their exponent and obtain \[ f(z) =\sum_{A\subset\mathbb N\,\text{finite}}z^{\sum_{a\in A}a} =\sum_{m=0}^{\infty}p(m)z^m. \]
3. If \(\zeta\) is a root of unity of odd order \(q\), then\[ |f(r\zeta)|\longrightarrow\infty \qquad (r\to1^-). \]
Since the product in 1 has no zeros in \(\Delta\), choose the analytic logarithm determined by \(\log f(0)=0\). Expanding \(\log(1+w)\) and using absolute convergence for \(|z|<1\) gives \[ \log f(z) =\sum_{n=1}^{\infty}\log(1+z^n) =\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\frac{z^k}{1-z^k}. \] Write \(r=e^{-s}\) with \(s>0\) and set \(z=r\zeta\). Split this series into terms with \(q\mid k\) and terms with \(q\nmid k\).
For \(k=q\ell\), since \(q\) is odd and \(\zeta^q=1\), the divisible part is \[ M(s) =\frac1q\sum_{\ell=1}^{\infty} \frac{(-1)^{\ell+1}}{\ell}\frac1{e^{q\ell s}-1}. \] Furthermore \[ 0<\frac{s}{e^{q\ell s}-1}\le\frac1{q\ell}. \] Dominated convergence therefore yields \[ \lim_{s\to0^+}sM(s) =\frac1{q^2}\sum_{\ell=1}^{\infty}\frac{(-1)^{\ell+1}}{\ell^2} =\frac{\pi^2}{12q^2}>0. \]
For the remaining terms, the finitely many numbers \(\zeta^j\), \(1\le j<q\), stay a positive distance from the radial segment ending at \(1\). Hence there is a constant \(c_q>0\) such that \[ |1-r^k\zeta^k|\ge c_q \qquad(q\nmid k,\ 0<r<1). \] Thus the nondivisible part \(N(s)\) satisfies \[ |N(s)| \le \frac1{c_q}\sum_{k=1}^{\infty}\frac{r^k}{k} =\frac1{c_q}\log\frac1{1-r} =O(\log(1/s)). \] Consequently \(sN(s)\to0\). We conclude that \[ \operatorname{Re}\log f(r\zeta) =\frac{\pi^2}{12q^2s}+o(1/s)\longrightarrow+\infty, \] and hence \(|f(r\zeta)|\to\infty\).
4. Roots of unity of odd order are dense in \(S^1\).
They are the points \(e^{2\pi i a/q}\) with \(q\) odd. Rationals with odd denominator are dense in \(\mathbb R/\mathbb Z\): given any real number and any \(\varepsilon>0\), choose an odd \(q\) so large that \(1/q<\varepsilon\) and then choose an integer \(a\) with \(|a/q-x|\le1/(2q)\). Passing through \(t\mapsto e^{2\pi it}\) proves the claim.
5. The unit circle is a natural boundary for \(f\). In particular, if\(g\) is analytic on a connected domain \(U\supset\Delta\) and \(g=f\) on\(\Delta\), then \(U=\Delta\).
Suppose instead that \(U\) properly contains \(\Delta\). Since an open connected subset of \(\mathbb C\) is path connected, there is a path in \(U\) from a point of \(\Delta\) to a point outside \(\overline\Delta\). The path meets \(S^1\) at some point \(\xi\in U\). Because \(U\) is open, some disk \(D(\xi,\varepsilon)\) lies in \(U\); its intersection with \(S^1\) contains an open arc.
By 4 choose an odd-order root of unity \(\zeta\) in that arc. Then \(\zeta\in U\), so analyticity of \(g\) makes \(g\) bounded on a sufficiently small closed disk about \(\zeta\). But for \(r<1\) sufficiently close to \(1\), \(r\zeta\) lies in that disk and also in \(\Delta\), hence \[ g(r\zeta)=f(r\zeta). \] This contradicts 3, which says \(|f(r\zeta)|\to\infty\). Therefore no such proper \(U\) exists, and \(U=\Delta\).