Prove that every Taylor coefficient at the origin of \[ f(z)=\frac{z}{e^z-1} \] is rational.
Solution
1. The singularity of \(f\) at \(0\) is removable, and its holomorphicextension satisfies\[ f(0)=1. \]
Write \[ e^z-1 = z \left( 1+\frac z{2!}+\frac{z^2}{3!}+\cdots \right). \] The factor in parentheses is holomorphic and equals \(1\) at \(0\), hence is nonzero near \(0\). Therefore \[ f(z) = \left( 1+\frac z{2!}+\frac{z^2}{3!}+\cdots \right)^{-1} \] extends holomorphically across \(0\) with value \(1\).
2. Write the Taylor expansion\[ f(z)=\sum_{n=0}^{\infty}a_nz^n \]near \(0\). Then\[ a_0=1 \]and, for every \(n\geq1\),\[ a_n = -\sum_{m=2}^{n+1} \frac{a_{n+1-m}}{m!}. \]
The identity \[ (e^z-1)f(z)=z \] holds near \(0\). Expanding, \[ \left( \sum_{m=1}^{\infty}\frac{z^m}{m!} \right) \left( \sum_{j=0}^{\infty}a_jz^j \right) = z. \] The coefficient of \(z\) gives \(a_0=1\). For \(n\geq1\), the coefficient of \(z^{n+1}\) is zero, so \[ \sum_{m=1}^{n+1}\frac{a_{n+1-m}}{m!}=0. \] The \(m=1\) term is \(a_n\), which gives the displayed recurrence.
3. Every coefficient \(a_n\) is rational.
Proceed by induction on \(n\). Step 2 gives \[ a_0=1\in\QQ. \] Assume \(a_0,\ldots,a_{n-1}\in\QQ\). The recurrence in step 2 expresses \(a_n\) as a finite sum of the earlier coefficients multiplied by the rational numbers \(1/m!\). Hence \(a_n\in\QQ\).
4. Q.E.D.
Step 3 proves that every Taylor coefficient of \(f\) at the origin is rational.