Let $d$ be a positive integer and $p(x)=1+x+x^2+\cdots+x^d$. Let $0<r<1$ and consider the product $$p(r^n x)p(r^{n-1}x)\cdots p(rx)=1+c_{1,n}x+c_{2,n}x^2+\cdots c_{nd,n}x^{nd},$$ where $c_{j,n}$ denotes the coefficient of $x^j$ in the above expansion. Writing out the product, we see by inspction $c_{1,n}=r+r^2+\cdots +r^{n}=r\frac{1-r^{n}}{1-r}$. In general, I think one can prove $$ c_{j,n}=\sum_{ j_1+j_2+\cdots +j_n=j} r^{j_1}r^{2j_2}\cdots r^{nj_n},$$ where the sum is taken over all nonnegative integer combinations of $j_i$ for which $0\leq j_i\leq d$ and $j_1+\cdots +j_n=j$.
Question: Is the doubly-indexed sequence $c_{j,n}$ bounded, meaning there exists some $M>0$ such that $c_{j,n}\leq M$ for all $j$ and $n$?
Let $c_{0,n}=1$ for all $n$. The following property may be useful: by considering the values that $j_n$ can take, namely $j_n=0,1,...,\min\{d,j\}$, we have \begin{align} c_{j,n}& =\sum_{ j_1+\cdots +j_{n-1}+0=j} r^{j_1}r^{2j_2}\cdots r^{(n-1)j_{n-1}} + \sum_{ j_1+\cdots +j_{n-1}+1=j} r^{j_1}r^{2j_2}\cdots r^{(n-1)j_{n-1}}r^{n}+\cdots +\sum_{ j_1+\cdots +j_{n-1}+\min\{d,j\}=j} r^{j_1}r^{2j_2}\cdots r^{(n-1)j_{n-1}} \\ & = c_{j,n-1} + r^n c_{j-1,n-1} + r^{2n}c_{j-2,n-1}+\cdots + c_{j-\min\{d,j\},n-1}r^{\min\{d,j\}n}\\ & \leq c_{j,n-1} + r^n c_{j-1,n-1} + r^{2n}c_{j-2,n-1}+\cdots + c_{0,n-1}r^{jn} \end{align}
Using this relationship $n-1$ times, I was able to show that if $d\geq 2$, then since $c_{2,1}=r^2$, \begin{align} c_{2,n}&=c_{2,1} + (r^n c_{1,n-1} + r^{n-1}c_{1,n-2}+\cdots+r^2 c_{1,1}) + (r^{2n}+r^{2(n-1)}+\cdots+r^{2(2)})\\ & = r^2\frac{1-r^{2n-1}}{1-r^2} + \sum_{i=1}^{n-1} r^{i+1}c_{1,i}. \end{align} As $c_{1,i}=r+r^2+\cdots+r^i=r\frac{1-r^i}{1-r}$, we have $$ \sum_{i=1}^{n-1} r^{i+1}c_{1,i} = \frac{r}{1-r} \sum_{i=1}^{n-1} (r^{i+1}-r^{2i+1})=\frac{r}{1-r}\left(r^2\frac{1-r^{n-1}}{1-r}-r^3\frac{1-r^{2n-3}}{1-r^2}\right).$$ If this is correct, then I have an exact formula for $c_{2,n}$, and can see in the limit that $$c_{2,n}\to \frac{r^2}{1-r^2}+\frac{r}{1-r} \left(\frac{r^2}{1-r}-\frac{r^3}{1-r^2}\right)=\frac{r^2}{(1-r)(1-r^2)}. $$ Numerical evidence suggests this is the correct limit of $c_{2,n}$. Unfortunately, computing $c_{3,n}$ using the method above is quite painful, and I have to imagine this method might be too taxing to handle the general case $c_{j,n}$. Numerical evidence also suggests that $c_{3,n}< \frac{r^3}{(1-r)(1-r^2)(1-r^3)}$, but I have not been able to show that such a pattern is emerging.
Of course, I don't necessarily want to know a closed formula for $c_{j,n}$; I just want to know if this (double) sequence is bounded. Perhaps there is a more analytic method for estimating these coefficients using the fact that for $|x|<1$, $p(x)=\frac{1-x^{d+1}}{1-x}$, which makes $$ p(r^n x)p(r^{n-1}x)\cdots p(rx)=\frac{1-(r^{n}x)^{d+1}}{1-r^n x} \frac{1-(r^{n-1}x)^{d+1}}{1-r^{n-1} x} \cdots \frac{1-(rx)^{d+1}}{1-r x}.$$ I could not find a use for this relationship, but would be happy if it was beneficial.
I may have found the solution while driving home. It appears that \begin{align} c_{j,n}& \leq p(r)p(r^2)\cdots p(r^n)\\ & = (1+r+r^2+\cdots+r^d)(1+r^2+r^4+\cdots+r^{2d})\cdots(1+r^{n}+r^{2n}+\cdots +r^{dn}). \end{align} Indeed, if $j_1+j_2+\cdots +j_n=j$ with $0\leq j_i\leq d$, then $r^{j_1}r^{2j_2}\cdots r^{nj_n}$ is a term in the expanded product above (take $r^{j_1}$ from $p(r)$, $r^{2j_2}$ from $p(r^2)$, and so on).
We claim the product $\prod_{n=1}^\infty p(r^n)$ converges. Using the fact that $p(r^j)=\frac{1-r^{j(d+1)}}{1-r^j}\leq \frac{1}{1-r^j}$, it suffices to show that $\prod_{n=1}^\infty \frac{1}{1-r^n}$ converges. The sum $\sum_{n=1}^\infty \log\left(\frac{1}{1-r^n}\right)=-\sum_{n=1}^\infty \log(1-r^n)$ converges absolutely since $\sum_{n=1}^\infty r^n$ converges absolutely. Thus $\prod_{n=1}^\infty p(r^n)$ converges, and $c_{j,n}\leq \prod_{n=1}^\infty p(r^n)$ for any $j$ between $0$ and $nd$.