Define \[ \zeta(x)=\sum_{n=1}^\infty\frac1{n^x}. \] Prove that \(\zeta(x)\) is defined and has continuous derivatives of all orders for \[ 1<x<\infty. \]
Solution
1. For every \(x>1\), the series\[ \zeta(x)=\sum_{n=1}^\infty n^{-x} \]converges.
For fixed \(x>1\), this is the convergent \(p\)-series with exponent \(p=x\). Thus \(\zeta\) is defined on \((1,\infty)\).
2. Let \(I=[\alpha,\beta]\subset(1,\infty)\) be compact. For everyinteger \(k\geq0\), the series\[ \sum_{n=1}^\infty (-\log n)^k n^{-x} \]converges uniformly for \(x\in I\).Here the factor \((-\log n)^0\) is understood as \(1\); for \(k\geq1\), the\(n=1\) term is \(0\).
For \(x\in I\), \[ \abs{(-\log n)^k n^{-x}} \leq (\log n)^k n^{-\alpha}. \] Set \[ \varepsilon\coloneqq\frac{\alpha-1}{2}>0. \] Since \[ (\log n)^k=o(n^\varepsilon), \] there is \(N\) such that, for \(n\geq N\), \[ (\log n)^k\leq n^\varepsilon. \] Therefore \[ (\log n)^k n^{-\alpha} \leq n^{-(\alpha-\varepsilon)} = n^{-(\alpha+1)/2}. \] Because \((\alpha+1)/2>1\), the last series converges. The Weierstrass M-test proves uniform convergence on \(I\).
3. For every integer \(k\geq0\) and every \(x>1\),\[ \zeta^{(k)}(x) = \sum_{n=1}^\infty (-\log n)^k n^{-x}. \]
For \[ f_n(x)\coloneqq n^{-x}=e^{-x\log n}, \] one has \[ f_n^{(k)}(x)=(-\log n)^k n^{-x}. \] Step 2 shows that, on the arbitrary compact interval \(I\), the series of \(k\)th derivatives converges uniformly for every \(k\geq0\). Starting with the uniformly convergent series from the case \(k=0\), apply the standard termwise-differentiation theorem successively: uniform convergence of \(\sum f_n^{(k+1)}\) permits differentiating the sum \(\sum f_n^{(k)}\). Induction on \(k\) gives the displayed formula throughout \(I\).
4. Every derivative \(\zeta^{(k)}\) is continuous on \((1,\infty)\).
Fix \(x_0>1\) and choose a compact interval \[ I=[\alpha,\beta]\subset(1,\infty) \] with \(x_0\) in its interior. By step 3, \(\zeta^{(k)}\) on \(I\) is the uniform limit of the continuous functions \[ \sum_{n=1}^N(-\log n)^k n^{-x}. \] Hence \(\zeta^{(k)}\) is continuous at \(x_0\). Since \(x_0\) and \(k\) were arbitrary, every derivative is continuous on \((1,\infty)\).
5. Therefore\[ \boxed{\zeta\in C^\infty((1,\infty))}. \]
Step 1 proves that \(\zeta\) is defined on \((1,\infty)\), while steps 3 and 4 prove existence and continuity of derivatives of every order there.
6. Q.E.D.
Step 5 is the required conclusion.