In my research in Quantum Field Theory, I have encountered two questions that involve partial Bell polynomials:


1. Let $u$ and $x\_i$ be indeterminates. I have checked that the following conjectured identity


$$
\sum \_{m=0}^\infty u^m \frac{(m+1)!}{(2m+2)!} B\_{2m+2,m+2}(x\_1,\dots,x\_{m+1})=
\frac{1}{2} \left( \sum\_{n=0}^\infty u^n \frac{n!}{(2n+1)!} B\_{2n+1,n+1}(x\_1,\dots, x\_{n+1}) \right)^2
$$


holds for the first orders in $u$. Is this identity a known result? A proof or any reference would be very welcome.


2. I would like to find the radius of convergence of a similar infinite sum, in the particular case when $x\_j = j^{j-2}$, or at least prove that it is different from zero. Namely, I would like to discuss the convergence of


$$ 
f(u) = \sum\_{m=1}^\infty u^m \frac{(m-1)!}{(2m)!} B\_{2m,m+1}(1^{1-2},2^{2-2},3^{3-2},\dots,m^{m-2})
$$


near $u=0$, but I don't know how to estimate the asymptotic growth of $B\_{2m,m+1}(1^{1-2},2^{2-2},3^{3-2},\dots,m^{m-2})$ for large $m$. Again, any help or reference would be welcome.



Two questions about Bell polynomials
With respect to the first question, I'm not sure about whether the given identity is known, but here is a proof. It also expresses the involved series in terms of reversion of the series $\frac{t}{g(t)}$ as defined below.


It is convenient to express the given identity in terms of [ordinary Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials#Ordinary_Bell_polynomials):
$$\hat{B}\_{n,k}(\frac{x\_1}{1!},\frac{x\_2}{2!},\ldots,\frac{x\_{n-k+1}}{(n-k+1)!}) = \frac{k!}{n!}B\_{n,k}(x\_1,x\_2,\ldots,x\_{n-k+1})$$
and omitting the indeterminates $x\_j$ and multiplying by $u^2$ as
$$\sum\_{m=0}^{\infty} \frac{u^{m+2}}{m+2} \hat{B}\_{2m+2,m+2} = \frac{1}{2}\left(\sum\_{n=0}^{\infty} \frac{u^{n+1}}{n+1} \hat{B}\_{2n+1,n+1} \right)^2.$$


From the [generating function for $\hat{B}\_{n,k}$](https://en.wikipedia.org/wiki/Bell_polynomials#Generating_function) it follows that
$$\hat{B}\_{2m+2,m+2} 
= [t^{2m+2}]\ \left(\sum\_{j\geq 1} \frac{x\_j}{j!}t^j\right)^{m+2} = [t^{m}]\ g(t)^{m+2}$$
and
$$\hat{B}\_{2n+1,n+1} 
= [t^{2n+1}]\ \left(\sum\_{j\geq 1} \frac{x\_j}{j!}t^j\right)^{n+1} = [t^n]\ g(t)^{n+1},$$
where $g(t):=\sum\_{j\geq 1} x\_j\frac{t^{j-1}}{j!}$.


Using [Lagrange–Bürmann formula](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula), we get
$$\sum\_{m\geq 0} \hat{B}\_{2m+2,m+2} t^{m+1} = \frac{w(t)w'(t)}t,$$
$$\sum\_{n\geq 0} \hat{B}\_{2n+1,n+1} t^n = w'(t),$$
where function $w(t)$ satisfies the functional equation: $w(t)=tg(w(t))$.


Correspondingly,
$$\sum\_{m\geq 0} \hat{B}\_{2m+2,m+2} \frac{u^{m+2}}{m+2} = \int\_0^u w(t) w'(t)\ {\rm d}t = \frac12 w(u)^2,$$
$$\sum\_{n\geq 0} \hat{B}\_{2n+1,n+1} \frac{u^{n+1}}{n+1} = \int\_0^u w'(t) {\rm d}t = w(u),$$
from where the required identity follows instantly.




---


As for the second question, we first notice that $f'(u) = \frac12 w(u)^2$. Also, when $x\_j=j^{j-2}$, we have $$g(t)=\sum\_{j\geq 1} j^{j-2}\frac{t^{j-1}}{j!},$$
which can be expressed in terms of Lambert W function as
$$g(t) = \frac{1}{2t}\left(1-(1+W\_0(-t))^2\right).$$
Then convergence of $w(t)$ can be imposed from viewing it as a [series reversion](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem) of $\frac{t}{g(t)}$.


