Why is this sum involving the Riemann Zeta function convergent for $\Re(s)>-1$?

Question

In this Mathworld article, the following function is defined (equation $[12]$):

$$ F(s) = (1-2^{1-s})\zeta(s), $$

where $\zeta(s)$ is the Riemann Zeta function defined by the usual infinite sum formula $\sum_{n\geq1} n^{-s}$. Then it is claimed that $F(s)$ converges for $\Re(s)>-1$. But I don't see how this can be true, given that $\sum_{n\geq1} n$ clearly diverges.

How can this claim be proven?

Answer 1

For $\Re(s) > 1$ $$(1-2^{1-s}) \zeta(s) = \sum_{n=1}^\infty n^{-s}-2 \sum_{n=1}^\infty (2n)^{-s}= \sum_{n=1}^\infty (-1)^{n+1} n^{-s}\\ = \sum_{n=1}^\infty (\sum_{m=1}^n (-1)^{m+1}) (n^{-s}-(n+1)^{-s}) = \sum_{n=1}^\infty (\frac12+\frac12 (-1)^{n+1}) (n^{-s}-(n+1)^{-s})\\ = \frac12 +\frac12 \sum_{n=1}^\infty (-1)^{n+1} (n^{-s}-(n+1)^{-s}) \qquad (1)$$

Because $$n^{-s}-(n+1)^{-s} = \int_n^{n+1} s t^{-s-1} dt =s n^{-s-1}+\int_n^{n+1} s (t^{-s-1}-n^{-s-1}) dt\\= s n^{-s-1}-s(s+1)\int_n^{n+1} \int_n^t u^{-s-2}du dt$$ $(1)$ converges and is analytic for $\Re(s) > -1$, thus it must be the analytic continuation of $(1-2^{1-s}) \zeta(s)$.

Answer 2

Usually, $\zeta(s)$ denotes the analytic continuation of $$\sum_{n=1}^\infty n^{-s},$$ which means $\zeta(s)$ is holomorphic in $\mathbb C\setminus\{1\}$. The region of definition has already been extended. In other words, $\zeta:\mathbb C\setminus\{1\}\to\mathbb C$.
$\zeta(s)$ does not equal $\sum_{n=1}^\infty n^{-s}$ if $s<1$, it is defined in $(-\infty,1)$ by using a different method (contour integration).
Note that $F(s)$ is not defined as $(1-2^{1-s})\sum_{n=1}^\infty n^{-s}$, it equals $(1-2^{1-s})\zeta(s)$, which gives a chance for $F(s)$ to converge when $\Re s\in(-1,1)$.