Define \[ f(x) = \sum_{n=1}^{\infty} \frac{1}{n^{x}}. \] Show that \(f\) converges to a differentiable function on \((1, \infty)\) and that \[ f'(x) =\sum_{n=1}^{\infty}\left(\frac{1}{n^{x}}\right)^{\prime}. \]
Hint: \[ \left(\frac{1}{n^{x}}\right)' = -\frac{1}{n^{x}} \ln n \]
Solution
Let \[ f_N(x)=\sum_{n=1}^N n^{-x}. \] Fix a compact interval \([a,b]\subset(1,\infty)\). For \(x\in[a,b]\), \[ |n^{-x}|\le n^{-a}, \] and \(\sum n^{-a}\) converges. Hence \(\sum n^{-x}\) converges uniformly on \([a,b]\) by the Weierstrass \(M\)-test.
Also \[ \frac{d}{dx}n^{-x}=-(\log n)n^{-x}. \] Choose \(\delta=(a-1)/2>0\). Since \(\log n=o(n^\delta)\), there is \(C>0\) such that \[ \log n\le Cn^\delta \] for every \(n\ge2\). Thus for \(x\in[a,b]\), \[ |(\log n)n^{-x}| \le Cn^{-(a-\delta)}. \] Because \[ a-\delta=\frac{a+1}{2}>1, \] the comparison series converges. Therefore \[ \sum_{n=1}^\infty -(\log n)n^{-x} \] converges uniformly on \([a,b]\).
The standard theorem on termwise differentiation of a series of \(C^1\) functions now applies: the original series converges at every point of \([a,b]\) (indeed uniformly there), and the derivative series converges uniformly. Hence its sum is differentiable on \([a,b]\) and \[ f'(x)=-\sum_{n=1}^\infty\frac{\log n}{n^x}. \] Since every \(x>1\) lies in such a compact interval, \[ \boxed{f'(x)=\sum_{n=1}^\infty\left(n^{-x}\right)'\qquad(x>1).} \]