Continuity of $\sum(-1)^n\left(\frac1{(n+1)^s} -\frac1{n^s}\right)$ (Dirichlet function continuation)

Question

How to show the continuity of $\sum(-1)^n\left(\dfrac{1}{(n+1)^s} -\dfrac{1}{n^s}\right)$ in $]-1,0]$ ?

(the continuity on $[0;+\infty[$ comes from the continuity of the Dirichlet's $\eta$ function.

Where $\sum(-1)^n\left(\dfrac{1}{(n+1)^s} -\dfrac{1}{n^s}\right)$ is a function serie of $s\in]-1;+\infty[$

Answer 1

The continuity follows (for example) from uniform convergence on $[-1+\delta,0]$ for every $\delta > 0$. To establish this, we use the facts that

1. for every fixed $s \in (-1,0]$ the sequence $(n+1)^{-s} - n^{-s}$ is monotonically decreasing and converges to $0$, and
2. for every fixed $n$, the function $s \mapsto (n+1)^{-s} - n^{-s}$ is decreasing on $(-1,0]$.

If these are established, we note that for every $s\in (-1,0]$

$$\Biggl\lvert \sum_{n = K}^M (-1)^n\bigl((n+1)^{-s} - n^{-s}\bigr)\Biggr\rvert \leqslant (K+1)^{-s} - K^{-s}$$

due to the alternating signs and fact 1, and by fact 2, we have consequently

$$\Biggl\lvert \sum_{n = K}^M (-1)^n\bigl((n+1)^{-s} - n^{-s}\bigr)\Biggr\rvert \leqslant (K+1)^{1-\delta} - K^{1-\delta} \xrightarrow{K \to \infty} 0$$

uniformly for all $s \in [-1+\delta,0]$. This means the sequence of partial sums

$$F_m(s) = \sum_{n = 1}^m (-1)^n\bigl((n+1)^{-s} - n^{-s}\bigr)$$

is uniformly convergent on $[-1+\delta,0]$.

It remains to prove facts 1 and 2. For the proof of the first, we consider for fixed $s\in (-1,0]$ the differentiable function $g \colon x \mapsto x^{-s}$. By the mean value theorem,

$$(n+1)^{-s} - n^{-s} = g(n+1) - g(n) = \bigl((n+1)-n\bigr) \cdot g'(n + \vartheta_n) = -\frac{s}{(n+\vartheta_n)^{1+s}}$$

for some $\vartheta_n \in (0,1)$. Since $n + \vartheta_n < n+1 < (n+1) + \vartheta_{n+1}$ and $0 < 1+s \leqslant 1$, it follows that $(n + \vartheta_n)^{1+s} < (n+1+\vartheta_{n+1})^{1+s}$, whence $g(n+2) - g(n+1) \leqslant g(n+1) - g(n)$, and

$$0 \leqslant g(n+1) - g(n) \leqslant \frac{\lvert s\rvert}{n^{1+s}} \xrightarrow{n\to\infty} 0.$$

These inequalities are strict for $s \neq 0$.

To prove the second, we differentiate with respect to $s$:

\begin{align} \frac{d}{ds}\bigl((n+1)^{-s} - n^{-s}\bigr) &= -(n+1)^{-s}\log (n+1) + n^{-s}\log n \\ &= -\underbrace{(n+1)^{-s}}_{> 0}\cdot \underbrace{\log \frac{n+1}{n}}_{> 0} - \bigl(\underbrace{(n+1)^{-s} - n^{-s}}_{\geqslant 0}\bigr)\underbrace{\log n}_{\geqslant 0} \\ & < 0. \end{align}