- Let
\[ F (x) = \sum_ {n = 0} ^ {\infty} p (n) x ^ {n} = \prod_ {n = 1} ^ {\infty} \frac {1}{1 - x ^ {n}} \]
be the generating function for the partitions. Show that
\[ \log F (x) \sim \frac {\pi^ {2}}{6 (1 - x)} \quad \mathrm{as} x \to 1, \mathrm{with} 0 < x < 1. \]
[Hint: Use log \(\textstyle F ( x ) = \sum \log ( 1 / ( 1 - x ^ { n } ) )\) and log \(\textstyle ( 1 / ( 1 - x ^ { n } ) ) = \sum ( 1 / m ) x ^ { n m }\) , so
\[ \log F (x) = \sum {\frac {1}{m}} {\frac {x ^ {m}}{1 - x ^ {m}}}. \]
Use also \(m x ^ { m - 1 } ( 1 - x ) < 1 - x ^ { m } < m ( 1 - x ) . ]\)
Solution
For \(0<x<1\), absolute convergence allows us to expand and rearrange: \[ \begin{aligned} \log F(x) &=\sum_{n=1}^\infty -\log(1-x^n)\\ &=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{x^{nm}}m =\sum_{m=1}^\infty\frac1m\frac{x^m}{1-x^m}. \end{aligned} \tag{1} \] The elementary inequalities from the hint are \[ m x^{m-1}(1-x)\le 1-x^m\le m(1-x). \] Applying them termwise to (1) yields \[ \frac{x^m}{m^2} \le (1-x)\frac1m\frac{x^m}{1-x^m} \le \frac{x}{m^2}. \] Summing over \(m\), \[ \sum_{m=1}^\infty\frac{x^m}{m^2} \le (1-x)\log F(x) \le x\sum_{m=1}^\infty\frac1{m^2}. \] As \(x\uparrow1\), dominated convergence gives \[ \sum_{m=1}^\infty\frac{x^m}{m^2}\longrightarrow\zeta(2)=\frac{\pi^2}{6}, \] and the right-hand side has the same limit. Hence \[ (1-x)\log F(x)\longrightarrow\frac{\pi^2}{6}, \] which is exactly \[ \boxed{\log F(x)\sim\frac{\pi^2}{6(1-x)}}. \]