Uniform convergence of $\sum_{m=1}^{\infty} (-1)^m \{{(m+1)^s - m^s} \}$ for $0 < \Re(s) < 1$

Question

This question has been asked in here and a critical part of the answer given is wrong, I believe. Read the rules of MSE and a repetition of this question is not against the policy and the reason for re-asking is that it may take quite a long time to reach 50 points to make a comment over there, so please do not close/delete this post...

My question: In the answer linked, $$\left((m+1)^s-m^s \right)=\frac s{m^{1-s}} +O\left(\frac1{m^{2-s}}\right)$$ is correct and the right hand side of $$f(s)=\sum_{m=1}^\infty (-1)^{m}\left((m+1)^s-m^s \right)$$ converges for $0<\text{Re}(s)<1$. However, the rearrangement of the series to the form of $$f(s)=\sum_{m=1}^\infty \left((2m+1)^s-2(2m)^s+(2m-1)^s\right)$$ is not correct since the alternative series convergences for $0<\text{Re}(s)<1$ and according to Rudin's book rearrangement is possible when absolute convergence holds and for $f(s)$ absolute convergence does not hold for $0<\text{Re}(s)<1$. So how $\sum_{m=1}^{\infty} (-1)^m \{{(m+1)^s - m^s} \}$ convergences uniformly for $0 < \text{Re}(s) < 1$?

Answer 1

Nice question, there are 2 important misunderstandings to point out:

1. This is not a rearrangement. Given a series $$\sum_{n\in\mathbb{N}}a_n$$ A rearrangement of it, is another series $$\sum_{n\in\mathbb{N}}a_{\phi(n)}$$ Where $\phi:\mathbb{N}\to\mathbb{N}$ is some permutation of the natural numbers, this does not have this form, rather it is a telescoping, i.e. we took the series $$\sum_{n\in\mathbb{N}}a_n$$ and rewrote it as $$\sum_{n\in\mathbb{N}}(a_{2n}+a_{2n-1})$$ You can check that this preserves limits and the convergence status of an arbitrary series.

2. Secondly, you ask, under the incorrect assumption that this is a rearrangement, the question:

So how does $\sum_{m=1}^{\infty} (-1)^m \{{(m+1)^s - m^s} \}$ converge uniformly for $0<\text{Re}(s)<1$ ?

This implies that you have misunderstood the theorem in Rudin, it does not state that a rearrangement preserves the limit of a series if and only if the series is absolutely convergent. In fact, given any convergent series, absolutely convergent or not, there will be a (non-trivial) rearrangement which does preserve the limit.