I'm trying to show that $\frac{\sqrt{x}}{\tanh(\sqrt{x})}$ is real analytic at $x=0$ (with the principal determination of $\sqrt x$).
Apparently (i.e. thanks to graphs with Maple) $n \mapsto |f^{(n)}(x)|$ is decreasing for all $x$ in, say, $(-1,1)$ ; this would give a uniform bound on $\sup_{(-1,1)} |f^{(n)}|$ and so analyticity.
Is there some litterature on such functions ? I mean, is there an intelligent way of showing such a property ? (Derivating by hand gives rather horrible computations where negative powers of $x$ annihilate one another but in a rather obscure way).
Start with the power series expansions of $\sinh(x)$ and $\cosh(x)$: $$ \begin{align} \sinh(x)&=x + x^3/6+x^5/120+x^7/5040+\cdots\\ \cosh(x)&=1+x^2/2+x^4/24+x^6/720+\cdots\\ \tanh(x)=\sinh(x)/\cosh(x)&=x-x^3/3+2x^3/15-17x^5/315+\cdots\\ \coth(x)&=x^{-1}+x/3-x^3/45+2x^5/945-\cdots\\ x\coth(x)&=1+x^2/3-x^4/45+\cdots\,, \end{align} $$ and now you substitute $\sqrt x$ for $x$, using the fact that your final series has all terms of even degree.