Series about coefficients of multiplicative inverse of power series

Question

Let be an integer $d\geqslant 2$ and a real number $L\in(0,1)$. I consider the following formal power series $$T(x) := 1-L\,\sum_{1\leqslant j < d} x^j =: \sum_{n\geqslant 0} {a_n}x^n$$ with $a_0=1$, $a_n=-L$ for $1\leqslant n \leqslant d-1$, and $a_n=0$ for $n\geqslant d$. The reciprocal $T(x)^{-1}$ can also be expressed as a formal power series $$ T(x)^{-1} = \sum_{n\geqslant 0} b_{n,d} x^n $$ with $b_{0,d}=1$ and for $n\geqslant 1$ $$ b_{n,d} = L\,\sum_{1\leqslant i \leqslant \min\{n,d-1\}} b_{n-i,d}.$$ Does the series $ \sum_{n\geqslant 0} b_{n,d}^2$ converges for any $d\geqslant 2$ ?

Thanks to Hexhist's shorter version in this discussion (I'm surprised that I found nothing with constant coefficients), and if I'm not wrong, I have obtained a nicely expression of $b_{n,d}$ (maybe there exists an easier one) : \begin{align*} b_{n,d} &= \sum_{\lambda\in P_n} (-1)^{|\lambda|} |\lambda|! \prod_{1\leqslant i \leqslant n} \frac{a_i^{m_i}}{m_i !} = \sum_{\substack{\lambda\in P_n \\ i\geqslant d \Rightarrow m_i=0}} \binom{|\lambda|}{m_1,\ldots,m_{d-1}}\, L^{|\lambda|}, \end{align*} where $P_n$ denotes the finite set of all partitions of $n$, $m_i=m_i(\lambda)$ is the number of parts of size equal to $i$ in the partition $\lambda$, and $|\lambda|$ is the length of the partition $\lambda=(\lambda_1,\ldots,\lambda_n)\in P_n$.

To answer to my question, I try to find an equivalent (as $n\to\infty$) or maybe an upper bound is sufficient: does the multinomial coefficient be maximal when the $m_i$ are closed of each other ? If $1\leqslant n<d$, then $b_{n,d}=L(L+1)^{n-1}$, but if $n\geqslant d$ I haven't found anything... The condition $L(d-1)<1$ might be appear. My question may be easier than this one. Thank you in advance.

Answer 1

If $d=2$, $b_{n,d}=L^n$ and thus $\sum_n b_{n,d}^2$ converges, so we can assume $d\geqslant 3$. Thanks to Greg Martin's comment, if the smallest zero $\lambda$ of $T(x)$ has modulus strictly larger than $1$, then the $b_{n,d}$ decay exponentially in $n$ and thus any of their moments still converges. If we use Lagrange's bounds on the reflected polynomial $$ x^{d-1}-L\sum_{0\leqslant i \leqslant d-2} x^i,$$ we obtain $$|\lambda| \geqslant \frac{1}{\max\{1,L(d-1)\}}.$$ Thus, if $L(d-1)\leqslant 1$, then $|\lambda|\geqslant 1$, and $L(d-1)=1 \Leftrightarrow 1$ is a root (and it's the only one with modulus $1$ - see Alex Ravsky's comment). Finally, if $L(d-1)< 1$, then $|\lambda|>1$ and $\sum_n b_{n,d}^2$ thus converges.

EDIT 1 : $L(d-1)> 1$

Since $T(0)=1>0$ and $T(1)=1-L(d-1)<0$, there exists a root in $(0,1)$ thanks to intermediate value theorem and thus $b_{n,d}\to\infty$.

EDIT 2 : $L(d-1)= 1$

In this case, $1$ is the unique pole of smallest modulus of $1/T(x)$ (thanks to $|\lambda|\geqslant 1$ and Alex Ravsky's comment), and has multiplicty $1$ because $$ T(x) = \frac{d-1-x-\cdots-x^{d-1}}{d-1} = \frac{(1-x)(x^{d-2}+2x^{d-3}+\cdots+(d-2)x+d-1)}{d-1}.$$ Therefore $b_{n,d}\to C$ for some constant $C$ (see this). Finally, we have (cf. this) $$ \lim\limits_{n\to\infty} b_{n,d} = \lim\limits_{x\to 1} \frac{1-x}{T(x)} = \frac{2}{d}.$$

ADDITIONAL REMARKS

We can consider the generating function for partitions with part of sizes from $1$ to $d-1$: $$ \prod_{1\leqslant i <d} \frac{1}{1-y^i},$$ and therefore we have $$ b_{n,d}=[x^n]\biggl(\prod_{1\leqslant i <d} \frac{1}{1-Lx^i}\biggr).$$ Moreover, we have obtained the limit $$ \lim\limits_{n\to\infty} \sum_{\substack{\lambda\in P_n \\ i\geqslant d \Rightarrow m_i=0}} \binom{|\lambda|}{m_1,\ldots,m_{d-1}}\, (d-1)^{-|\lambda|} = \frac{2}{d}. $$