Exact convergent Ising-chain high-temperature series

At zero field the larger transfer-matrix eigenvalue in [F1] is ${\lambda }_{+}=e^K+e^{-K}=2\cosh K$. Therefore the limiting pressure is $P(K)=\log (2\cosh K)=\log 2+\log (1+w)$ with $w=\cosh K-1\ge 0$.

Put $u=\mathrm{\mid }K\mathrm{\mid }\mathrm{/}2<1\mathrm{/}4$. If $u=0$, then $w=0$. If $u>0$, [F3] applied to $\sinh$ on $[0,u]$ gives $\sinh u=u\cosh \xi$ for some $0<\xi <u$; by [F4], $\cosh \xi <\cosh u$, while [F6] and the definition of $\cosh$ give $\cosh u<e^u$, so $\sinh u<u{e}^u$. The power-series definition of $\exp$ in [F2], with $j!\ge 1$, and [F5] give $e^u={\sum }_{j\ge 0}{u}^j\mathrm{/}j!\le {\sum }_{j\ge 0}{u}^j=(1-u{)}^{-1}<4\mathrm{/}3$. Thus $0\le w=2{\sinh }^{2}u<2(4u\mathrm{/}3{)}^2=8K^2\mathrm{/}9<2\mathrm{/}9<1\mathrm{/}4$ when $u>0$, and $w=0$ when $u=0$.

For $0\le t\le w<1$, the finite geometric identity gives ${\sum }_{j=0}^n(-t{)}^j=(1-(-t{)}^{n+1})\mathrm{/}(1+t)$; the remainder is bounded uniformly by $w^{n+1}\mathrm{/}(1-w)$. The pointwise sum is also $(1+t{)}^{-1}$ by [F5], so these same remainder bounds show uniform convergence on $[0,w]$.

Integrate the finite geometric identity from $0$ to $w$. The integral of its remainder has absolute value at most $w^{n+2}$, since $1+t\ge 1$ on $[0,w]$; hence these finite integrals converge to ${\int }_0^w(1+t{)}^{-1}dt=\log (1+w)$ by [F2]. Integrating each finite sum gives ${\sum }_{m=1}^{n+1}(-1{)}^{m+1}{w}^m\mathrm{/}m$, so passage to the limit yields $\log (1+w)={\sum }_{m\ge 1}(-1{)}^{m+1}{w}^m\mathrm{/}m$. If $\mathrm{\mid }K\mathrm{\mid }\le r<1\mathrm{/}2$ with $r\ge 0$, step 1.2 gives $0\le w\le 8r^2\mathrm{/}9<2\mathrm{/}9=:w_r<1$. The tail of the series of absolute values after index $M$ is at most ${\sum }_{m>M}{w}_r^m=w_r^{M+1}\mathrm{/}(1-w_r)\to 0$, uniformly in $K$, by [F5]. This proves absolute convergence and uniform convergence on every such compact interval, including $r=0$.