Cayley coefficient proof of the first Dahlquist barrier ($p\le s+1(s\text{ odd}),p\le s+2(s\text{ even})$)

For a real consistent zero-stable linear multistep method, transform its first characteristic polynomial by $z=(1+x)/(1-x)$ to $(1-x{)}^s\rho (z)=xP(x)$. The root condition for a multistep method puts all finite roots of $P$ in the closed left half-plane; real linear factors and conjugate quadratic factors consequently give nonnegative coefficients, after choosing $P(0)>0$. The order criterion requires a polynomial of degree at most $s$ to agree through degree $p-1$ with $P(x)x/(2\mathrm{arctanh}x)$. The latter quotient has strictly negative nonconstant even coefficients. Its first even coefficient beyond degree $s$ includes a strictly negative multiple of $P(0)$ and cannot vanish. This gives the stated bound.

Write $F(x)={\int }_0^1(1+x{)}^t(1-x{)}^{1-t}dt=x/\mathrm{arctanh}x$. Pairing $t$ and $1-t$ after differentiating twice givesThe power series of $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients, and the other factor has nonnegative even coefficients. Thus every even coefficient of $F^{\prime \prime }$ is negative. Since $F$ is even with $F(0)=1$, the displayed sign pattern follows. This supplies a short positivity argument for the first Dahlquist barrier.