Determine the radius of convergence of \(\sum_{n=1}^{\infty}a_nz^n\) when:
\(a_n=(\log n)^2\);
\(a_n=n!\);
\(a_n=\dfrac{n^2}{4^n+3n}\);
\(a_n=\dfrac{(n!)^3}{(3n)!}\);
for the hypergeometric series \[ F(\alpha,\beta,\gamma;z) = 1+\sum_{n=1}^{\infty} \frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\,\beta(\beta+1)\cdots(\beta+n-1)} {n!\,\gamma(\gamma+1)\cdots(\gamma+n-1)}z^n, \] where \(\alpha,\beta\in\mathbb C\) and \(\gamma\ne0,-1,-2,\ldots\);
for the Bessel function of positive integer order \(r\), \[ J_r(z) =\left(\frac z2\right)^r \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+r)!}\left(\frac z2\right)^{2n}. \]
Solution
1. In part (a), the radius of convergence is \(R=1\).
By the Cauchy-Hadamard formula, \[ R^{-1}=\limsup_{n\to\infty}|a_n|^{1/n}. \] Here \[ |a_n|^{1/n}=\exp\!\left(\frac{2\log\log n}{n}\right)\longrightarrow1, \] so \(R^{-1}=1\).
2. In part (b), the radius of convergence is \(R=0\).
For the nonzero coefficients \(a_n=n!\), \[ \left|\frac{a_{n+1}}{a_n}\right|=n+1\longrightarrow\infty. \] Thus for every \(z\ne0\), the ratio of successive absolute values of the terms is \((n+1)|z|\to\infty\), so the terms do not even tend to zero. Hence convergence occurs only at \(z=0\).
3. In part (c), the radius of convergence is \(R=4\).
One has \[ |a_n|^{1/n} = \frac{n^{2/n}}{(4^n+3n)^{1/n}}. \] Now \(n^{2/n}\to1\), while \[ (4^n+3n)^{1/n} =4\left(1+\frac{3n}{4^n}\right)^{1/n}\longrightarrow4. \] Therefore \(\lim |a_n|^{1/n}=1/4\), so Cauchy-Hadamard gives \(R=4\).
4. In part (d), the radius of convergence is \(R=27\).
The coefficient ratio is \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{(3n+1)(3n+2)(3n+3)} \longrightarrow\frac1{27}. \] Hence the ratio of successive absolute values of the terms \(a_nz^n\) tends to \(|z|/27\). The ratio test gives convergence for \(|z|<27\) and divergence for \(|z|>27\), so the radius is \(27\).
5. In part (e), if neither \(\alpha\) nor \(\beta\) is a nonpositive integer, the hypergeometric series has radius \(R=1\).
Let \[ c_n= \frac{(\alpha)_n(\beta)_n}{n!(\gamma)_n}, \] where \((q)_n=q(q+1)\cdots(q+n-1)\). Under the stated assumption on \(\alpha\) and \(\beta\), all \(c_n\) are nonzero, and the hypothesis on \(\gamma\) makes the denominator nonzero. Thus \[ \left|\frac{c_{n+1}}{c_n}\right| = \left|\frac{(n+\alpha)(n+\beta)}{(n+1)(n+\gamma)}\right| \longrightarrow1. \] The ratio test therefore gives convergence for \(|z|<1\) and divergence for \(|z|>1\).
6. If \(\alpha\) or \(\beta\) is a nonpositive integer, the hypergeometric series terminates and has radius \(R=\infty\).
If, for example, \(\alpha=-m\) with \(m\in\mathbb Z_{\ge0}\), then \((\alpha)_n=0\) for every \(n\ge m+1\). Hence only finitely many coefficients are nonzero, so \(F\) is a polynomial. The same argument applies to \(\beta\).
7. In part (f), the Bessel series has radius \(R=\infty\).
Set \[ c_n=\frac{(-1)^n}{n!(n+r)!}. \] As a power series in \(w=z^2/4\), its coefficient ratio satisfies \[ \left|\frac{c_{n+1}}{c_n}\right| = \frac1{(n+1)(n+r+1)}\longrightarrow0. \] Hence \(\sum c_nw^n\) converges for every \(w\in\mathbb C\). Substituting \(w=z^2/4\) therefore gives convergence for every \(z\in\mathbb C\), and multiplication by the polynomial factor \((z/2)^r\) does not change this. Thus \(J_r\) is entire and its radius of convergence is infinite.