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Put $F(s)=(s-1)\zeta (s)$. The fractional-part continuation formula for the Riemann zeta function gives $|F(s)|\ll (|s|+1{)}^2$ for $\mathrm{Re}s\ge 1/4$. At $2+ij$, the reciprocal Euler product bounds $|\zeta (2+ij)|$ below by $1/\zeta (2)$. Applying the Jensen zero-count bound with outer radius $7/4$ and inner radius $8/5$ gives $O(\log (|j|+2))$ zeros in the latter disk. Those disks cover the half-strip $1/2\le \mathrm{Re}s\le 1$. Reflecting zeros using the functional equation of the Riemann zeta function completes the $O(T\log T)$ bound without requiring left-half-plane growth estimates.

For $\Re s>1$, the Riemann zeta function is the absolutely convergent Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$. Apply Abel summation to the tail, with counting function $\lfloor w\rfloor -\lfloor x\rfloor$. The upper boundary term tends to zero, givingSubstitute $\lfloor w\rfloor =w-\{w\}$ and integrate the $w$ term. This proves the fractional-part continuation formula for the Riemann zeta functionThe endpoint convention is valid whether or not $x$ is an integer. Since $0\le \{w\}<1$, the last integral converges locally uniformly for $\Re s>0$, including after differentiation on compact subsets. It therefore defines a holomorphic function there. The other terms are entire except for $x^{1-s}/(s-1)$. By the identity theorem for holomorphic functions, the formula supplies a meromorphic continuation with exactly one pole in $\Re s>0$: a simple pole at $s=1$ of residue one.

For $1<\sigma \le 2$, the three-four-one inequality for Euler products gives $|1/\zeta (\sigma +it)|\le \zeta (\sigma {)}^{3/4}|\zeta (\sigma +2it){|}^{1/4}$. For $|t|\ge 1$, the fractional-part continuation formula for the Riemann zeta function, truncated at $H\asymp 2+|t|$, gives ${\zeta }^{(j)}(\sigma +it)=O_j({\log }^{j+1}(2+|t|))$. Thus the reciprocal bound even holds with exponent $1/4$ instead of $1/2$ in this range. On bounded-height compact sets, three-four-one product proof of zeta boundary nonvanishing and removal of the reciprocal's zero at the pole give uniform bounds. Differentiating $1/\zeta$ repeatedly yieldsThe range restriction matters: $1/\zeta (\sigma )\to 1$ as $\sigma \to \infty$, so the first displayed bound cannot hold uniformly for all $\sigma >1$. A global version replaces $(\sigma -1{)}^{-3/4}$ with $1+(\sigma -1{)}^{-3/4}$.