Determine all \(z\in\mathbb C\) for which the power series \[ \sum_{n=1}^{\infty}\frac{z^n}{n^{\log n}} \] and every term-by-term derivative of this series converge absolutely.
Solution
For an integer \(m\geq0\), let \(S_m(z)\) denote the series obtained by differentiating term by term \(m\) times. Thus \[ S_m(z) = \sum_{n=\max\{1,m\}}^{\infty} \frac{n(n-1)\cdots(n-m+1)}{n^{\log n}}z^{n-m}, \] where for \(m=0\) the numerator is understood to be \(1\).
1. One has\[ \lim_{n\to\infty} \left(\frac{1}{n^{\log n}}\right)^{1/n} =1. \]
Since \[ n^{\log n} = e^{(\log n)^2}, \] we have \[ \left(\frac{1}{n^{\log n}}\right)^{1/n} = e^{-(\log n)^2/n}. \] Because \[ \frac{(\log n)^2}{n}\longrightarrow0, \] the displayed quantity tends to \(1\).
2. If \(\abs{z}>1\), then the original series does not converge.
For its \(n\)th term, \[ u_n=\frac{z^n}{n^{\log n}}, \] step 1 gives \[ \abs{u_n}^{1/n} = \abs{z} \left(\frac{1}{n^{\log n}}\right)^{1/n} \longrightarrow \abs{z}>1. \] Hence \(\abs{u_n}\) does not tend to zero, so the series cannot converge.
3. Fix \(m\geq0\). For all \(n\geq\max\{1,m\}\),\[ n(n-1)\cdots(n-m+1)\leq n^m. \]
Each of the \(m\) factors on the left is at most \(n\).
4. If \(\abs{z}<1\), then \(S_m(z)\) converges absolutely for every\(m\geq0\).
Fix \(m\geq0\) and put \(r=\abs{z}<1\). By step 3, \[ \left| \frac{n(n-1)\cdots(n-m+1)}{n^{\log n}}z^{n-m} \right| \leq n^m r^{n-m}, \] because \(n^{\log n}\geq1\) for \(n\geq1\). The series \[ \sum_{n=\max\{1,m\}}^{\infty}n^m r^{n-m} \] converges by the ratio test, since the ratio of consecutive terms tends to \(r<1\). Therefore \(S_m(z)\) converges absolutely.
5. If \(\abs{z}=1\), then \(S_m(z)\) converges absolutely for every\(m\geq0\).
Fix \(m\geq0\). By step 3 and \(\abs{z}=1\), \[ \left| \frac{n(n-1)\cdots(n-m+1)}{n^{\log n}}z^{n-m} \right| \leq n^{m-\log n}. \] Since \(\log n\to\infty\), there exists \(N\) such that \[ \log n\geq m+2 \] for every \(n\geq N\). Hence for \(n\geq N\), \[ n^{m-\log n}\leq n^{-2}. \] The tail is therefore dominated by the convergent series \[ \sum_{n=N}^{\infty}\frac1{n^2}, \] so \(S_m(z)\) converges absolutely.
6. The required set is\[ \boxed{\{z\in\CC:\abs{z}\leq1\}}. \]
Steps 4 and 5 show that the original series and every term-by-term derivative converge absolutely whenever \(\abs{z}\leq1\). Step 2 excludes every \(z\) with \(\abs{z}>1\).
7. Q.E.D.
Step 6 gives exactly the requested set of complex numbers.