Summation Formula for Tangent/Secant Numbers

Solution 1:

The paper Combinatorial aspects of continued fractions by P. Flajolet answers (affirmatively) all these questions. The following may help explain or simplify the arguments.

We treat a continued fraction like $f_N(x)$ as a formal power series, not as a function (because of convergence issues). That is, for a sequence $(a_1,a_2,a_3,\ldots)$ of, say, complex numbers, $$f(x)=\cfrac{1}{1-\cfrac{a_1 x}{1-\cfrac{a_2 x}{1-\cfrac{a_3 x}{\ldots}}}}:=\lim_{n\to\infty}\cfrac{1}{1-\cfrac{a_1 x}{1-\cfrac{\ldots}{1-\cfrac{a_n x}{1}}}}\tag{1}\label{cfracdef}$$ with $\big(1-g(x)\big)^{-1}:=\sum_{n=0}^{\infty}\big(g(x)\big)^n$ for $g(x)=\sum_{n=\color{red}{1}}^{\infty}g_n x^n$, and all limits understood in the conventional topology. In this setup, we have (see Corollary 2 in the paper) $$f(x)=1+\sum_{n=1}^{\infty}f_n x^n,\qquad f_n=\sum_{k_1=1}^{\color{gray}{0+}1}\sum_{k_2=1}^{k_1+1}\ldots\sum_{k_n=1}^{k_{n-1}+1}a_{k_1}a_{k_2}\ldots a_{k_n}.\tag{2}\label{cfpowers}$$ The paper gives it as a consequence of a more general treatment of continued fractions in a non-commutative setting (Theorem 1 there, with a "language-theoretic" proof that looks easy to read).

Put $k_{\ell}=\ell-m_{\ell}$ for $1\leqslant\ell\leqslant n$ here; then $f_n=\displaystyle\sum_{(m_1,\ldots,m_n)\in\mathscr{M}_n}\prod_{\ell=1}^{n}a_{\ell-m_{\ell}}$, where $$\mathscr{M}_n:=\left\{ (m_1,\ldots,m_n)\ \middle|\ \begin{array}{c}0\leqslant m_1\leqslant\ldots\leqslant m_n,\\ m_{\ell}\leqslant\ell-1\ (1\leqslant\ell\leqslant n)\end{array} \right\}.$$ Now if $\mathscr{N}_n:=\{ (m_1,\ldots,m_n) \mid 0\leqslant m_1\leqslant\ldots\leqslant m_n\leqslant n\}$, then $$\sum_{(m_1,\ldots,m_n)\in\mathscr{M}_n}\prod_{\ell=1}^{n}(\ell-m_{\ell})^N=\sum_{(m_1,\ldots,m_n)\in\mathscr{N}_n}\prod_{\ell=1}^{n}(\ell-m_{\ell})^N,$$ because each $(m_1,\ldots,m_n)\in\mathscr{N}_n\setminus\mathscr{M}_n$ has an $\ell$ with $m_{\ell}\geqslant\ell$ and hence with $m_{\ell}=\ell$.

This settles your question about the link between $\widehat{S}^{(N)}_{2n}\color{gray}{\text{ (why $2n$?)}}$ and $f_N$. The continued fractions for the secant and tangent numbers are also given in the paper (by Theorem 3B and Theorem 3A, respectively), but the combinatorial argument used is fairly hard to grasp. An alternative (attributed to Stieltjes) is to see that if the limit $f(x)$ in $\eqref{cfracdef}$ exists for any sufficiently small $x>0$, and if $f(x)$ admits an asymptotic power series in $x$, then this series is necessarily of the form $\eqref{cfpowers}$. Then, the secant numbers are found in the $s\to+\infty$ asymptotics of $$\cfrac{1}{1+\cfrac{1^2/s^2}{1+\cfrac{2^2/s^2}{1+\ldots}}}=2s\int_0^1\frac{x^s\,dx}{1+x^2}=s\int_0^\infty\frac{e^{-sx}\,dx}{\cosh x}$$ (using Watson's lemma), and similarly with the tangent numbers.