Let \(f\) be a meromorphic function on \(\CC\) which is analytic in a neighborhood of \(0\). Let its Maclaurin series be
\[ \sum_{k=0}^{\infty}a_kz^k \]
with all \(a_k\geqslant0\). Suppose there is a pole of modulus \(r>0\) and no pole has modulus \(<r\). Prove there is a pole at \(z=r\).
Solution
1. The Maclaurin series \(\sum_n a_nz^n\) has radius of convergenceexactly \(r\), and \(f\) is analytic on \(D(0,r)\).
The function \(f\) has no pole in \(D(0,r)\), so it is analytic there and its Maclaurin series converges on \(D(0,r)\). The series cannot converge on a larger disk, because its sum would then be analytic near the pole of modulus \(r\).
2. \(f\) is not analytic at \(r\).
2.1. Suppose, toward a contradiction, that \(f\) is analytic on \(D(r,\delta)\)for some \(\delta>0\). Fix \(\varepsilon\) with \(0<3\varepsilon<\delta\) andput \(x=r-\varepsilon\). Then \(f\) is analytic on \(D(x,2\varepsilon)\).
It suffices to show \[ D(x,2\varepsilon)\subset D(0,r)\cup D(r,\delta), \] since \(f\) is analytic on both disks by step 1 and the assumption. If \(z\in D(x,2\varepsilon)\) and \(\abs{z}<r\), then \(z\in D(0,r)\). Otherwise, \[ \abs{z-r}\le \abs{z-x}+\abs{x-r}<2\varepsilon+\varepsilon=3\varepsilon<\delta, \] so \(z\in D(r,\delta)\).
2.2. For every \(k\ge0\),\[ \frac{f^{(k)}(x)}{k!} =\sum_{n=k}^\infty \binom nk a_n x^{n-k}\ge0. \]
Since \(0<x<r\), the Maclaurin series may be differentiated term by term at \(x\), which gives the displayed formula. Every summand is nonnegative because \(a_n\ge0\) and \(x>0\).
2.3. The Maclaurin series converges at \(r+\tfrac12\varepsilon\).
Put \(h=\tfrac32\varepsilon\). Then \(0<h<2\varepsilon\), so by step 1 the Taylor series of \(f\) about \(x\) converges at \(x+h=r+\tfrac12\varepsilon\). By step 2 all terms of the double series below are nonnegative, so Tonelli’s theorem for series permits rearrangement: \[ \begin{aligned} f(x+h) &=\sum_{k=0}^\infty \frac{f^{(k)}(x)}{k!}h^k\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty \binom nk a_n x^{n-k}h^k\\ &=\sum_{n=0}^\infty a_n \sum_{k=0}^n\binom nk x^{n-k}h^k\\ &=\sum_{n=0}^\infty a_n(x+h)^n. \end{aligned} \] The left-hand side is finite.
2.4. Q.E.D.
Step 3 gives convergence of the Maclaurin series at the real point \(r+\tfrac12\varepsilon>r\). A power series converging at a point converges absolutely on the open disk of that radius, so its radius of convergence exceeds \(r\), contradicting step 1.
3. \(z=r\) is a pole of \(f\).
Since \(f\) is meromorphic on \(\CC\), every point is either a point of analyticity or a pole. Step 2 excludes the first case at \(r\).
4. Q.E.D.
Step 3 is the required conclusion.