Detailed proof of $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$

Solution 1:

Start with the defining series, convergent for $\Re(s)>1$: $$ \sum_{n=1}^\infty \frac{1}{n^s} = \sum_{n=1}^\infty \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \mathrm{e}^{-n t} \mathrm{d} t = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \frac{\mathrm{e}^{-t}}{1- \mathrm{e}^{-t}} \mathrm{d} t $$ Also notice that, for $\Re(s)>1$, $$ \frac{1}{s-1} = \frac{\Gamma(s-1)}{\Gamma(s)} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-2} \mathrm{e}^{-t} \mathrm{d} t $$ Now subtracting we have: $$ \zeta(s) - \frac{1}{s-1} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \mathrm{e}^{-t} \left( \frac{1}{1- \mathrm{e}^{-t}} - \frac{1}{t} \right) \mathrm{d} t $$ The integral on the right-hand side converges now for $\Re(s)>0$, since at the lower integration bound $\frac{1}{1-\exp(-t)} - \frac{1}{t} = \frac{1}{2} + \mathcal{o}(1)$.

Solution 2:

Trying to address what I think was troubling the OP.

Fix constants $a,b$ such that $0<a<b$. Consider the set $$L(a,b)=\{x+yi\in\mathbb{C}\mid a<x, |x+yi|<b\}.$$ The functions $g_n(z)$ are holomorphic in the set $L(a,b)$. Item $(ii)$ means that for all $z\in L(a,b)$ and all positive integers $n$ we have $$ |g_n(z)|\le\frac{b}{n^{1+a}}. $$ Therefore, by Weierstrass' M-test (note that here the lower bound on the exponent, $1+a$, does not depend on $z$, only on the set $L(a,b)$), the series $f(z)=\sum_{n=1}^\infty g_n(z)$ converges uniformly in the set $L(a,b)$ and therefore gives a holomorphic function on $L(a,b)$.

The right half-plane is the union of the sets $L(a,b)$, so $f(z)$ is holomorphic there.