For which \(z\in\CC\) does the series \[ \sum_{n=0}^\infty\left(\frac{z^n}{n!}+\frac{n^2}{z^n}\right) \] converge?
Solution
1. The series\[ \sum_{n=0}^\infty\frac{z^n}{n!} \]converges absolutely for every \(z\in\CC\).
For fixed \(z\), the ratio of successive absolute values is \[ \frac{|z|^{n+1}/(n+1)!}{|z|^n/n!} = \frac{|z|}{n+1} \longrightarrow0. \] The ratio test therefore gives absolute convergence.
2. If \(|z|>1\), then\[ \sum_{n=1}^\infty\frac{n^2}{z^n} \]converges absolutely.
For \[ b_n=\frac{n^2}{z^n}, \] one has \[ \frac{|b_{n+1}|}{|b_n|} = \left(\frac{n+1}{n}\right)^2\frac1{|z|} \longrightarrow \frac1{|z|}<1. \] The ratio test gives absolute convergence.
3. If \(|z|>1\), the series in the problem converges absolutely.
It is the termwise sum of the two absolutely convergent series from steps 1 and 2; the \(n=0\) term of the second series is \(0\).
4. If \(0<|z|\le1\), the summand in the problem does not tend to \(0\).
1. Hence\[ \left| \frac{z^n}{n!}+\frac{n^2}{z^n} \right| \ge \frac{n^2}{|z|^n}-\frac{|z|^n}{n!} \longrightarrow\infty. \]In particular, the summand does not tend to zero, so the series diverges.
5. The series is not defined at \(z=0\).
For every \(n\ge1\), the term \(n^2/z^n\) has zero denominator when \(z=0\).
6. Therefore the convergence set is\[ \boxed{\{z\in\CC:|z|>1\}.} \]
Step 3 proves convergence throughout this set, while steps <1>4–<1>5 exclude every \(z\) outside it.
7. Q.E.D.
Step 6 is the requested classification.