Coefficient of $\frac{1}{1-x-x^2-..-x^{d-1}}$ and its asymptotic 
Let $C_n$ be the set of all compositions of n such that each letter is one of ${1,2,\ldots,d-1}$.


A. Calculate the generating  function of the number of compositions in $C_n$.


B. For each composition $\pi \in C_n$ define  $f(\pi)$ as the number of letters that equal $1$.
Find yhe generating function of $\sum_{n\geq 0} \sum_{\pi \in C_n}  q^{f(\pi)} x^n$.


C.Find the expectation of the number of letters that equal 1 (1's appearances) in these compositions, and analyze it asymptotically.

*Generally: Composition of $n$: a sequence of numbers in $N$ such that their sum equals $n$. For example:
$C_0={\epsilon}$ , $C_1={1}$, $C_2={11,2}$ , $C_3={111,12,21,3}$.
In this question, a composition of a number $n$ , consists of numbers from the set {$1,2,\ldots, d-1$} where $d$ is a given number.
This is what I did:
A. Every compisition of n = empty compisition $\cup$ (1 / every compisition) $\cup$ (2 / every compisition) $\cup$....$\cup$((d-1) / every compisition).
so if x counts the length of a number, then the  generating function $C(x)$ for the problem satisfy:
$C(x)=1+xC(x)+x^2C(x)+\cdots+x^{d-1}C(x)$
(1 for the empty compisition)
$C(x)=1+C(x)(x+x^2+\cdots+x^{d-1})$
Therefore, $C(x)=\frac{1-x}{1-2x+x^d}=\frac{1}{1-x-x^2-...-x^{d-1}}$
B.Let $q$ counts the appearances of 1 and $x$ as we mentioned counts the length of a number.
Then:
$C(x)=1+qxC(x,q)+x^2C(x,q)+\cdots+x^{d-1}C(x,q)$
$C(x,q)=1+C(x,q)(qx+x^2+\cdots+x^{d-1})$
So, $C(x,q)=\frac{1-x}{x^d+(q-1)x^2-(q+1)x+1}$
C. We have to calculate $[x^n]C(x,1)=[x^n] C(x)$ and $[x^n] ([(d/dq) C(x,q)]_{q=1})=\frac{x}{(1-x-x^2-...-x^{d-1})^2}=xC(x)^2$ so the expectation will be the proportion of them.
But I could not see how the coefficients can be calculated!
I did not succeed to connect it to Fibonacci series which I know (the case of d=2).
I would be thankful for any help to show part C.
Thanks in advance
 A: The analysis of this sort of generating function's asymptotics is an exercise in partial fraction decomposition. Suppose we're trying to find the asymptotics of $P_n$ whose generating function $\sum_i P_ix^i$ is $\dfrac1{P(x)}$. If we can factor the polynomial $P(x)$ as $(x-r_1)(x-r_2)\cdots(x-r_n)$ (and all the roots are distinct), then we can write $\dfrac1{P(x)}$ as $\sum_i\dfrac{m_i}{x-r_i}$ for some constants $m_i$ (more on this later). We can then go the other direction, using the geometric series to write $P_n$ as $\sum_i \frac{m_i}{r_i}(r_i^{-1})^n$. If $r_1$ is the smallest root in absolute value, then this implies that $P_n=\frac{m_1}{r_1}(r_1^{-1})^n(1+O(\epsilon^n))$ for some $\epsilon\lt 1$. (There are very minor complications when the smallest roots are complex conjugates but that's not relevant here.)
In this case, we know that the polynomial $P(x)=1-(x+x^2+\ldots+x^{d-1})$ is positive at $x=0$ and negative at $x=1$ (as long as $d\gt 2$; if $d=2$ the analysis is trivial), so it has a root in $(0,1)$. What's more, since $P'(x)$ is negative on this interval, it has a single, simple root $\rho_{d-1}$ in the interval. (And in fact, all roots of $P()$ are simple since $\gcd(P, P')=1$.) Using Rouché's Theorem applied to $(1-x)P(x) = 1-2x+x^d$, we can show that there are no other roots of $P(x)$ in the disc $x\lt 1$, so this root dominates the asymptotics; we have that $C_n=C(\xi_{d-1})^n$ for some constant $C$, where $\xi_{d-1}=\rho_{d-1}^{-1}$. Note that we can say this even without being able to write an explicit formula for $\xi_{d-1}$ (except in the first few cases, such as the Fibonacci numbers).
The next piece of the puzzle is to compute $C$; fortunately, there's a standard tool for this, the residue method — I'll leave the details of this to you.
Also, one small additional note: if you're curious, you can also examine the asymptotics of $\xi_d$ itself — specifically, the asymptotics as $d$ goes to infinity. It should be straightforward to find the limiting value (hint: what $x$ has $x+x^2+x^3+\ldots=1$?) but the rate of approach is a really nice problem.
A similar analysis can be done in principle for $[x^n]\left(xC(x)^2\right)$ $=[x^{n-1}]\left(C(x)^2\right)$, but it's substantially complicated by the fact that the root is no longer simple. Fortunately, there's another way of finding the asymptotics: you can use the equation for the coefficients of $C(x)^2$ in terms of the $C_n$ and then applying the asymptotic analysis of the $C_n$ to those terms. (You should find that each of the summands in the equation for the coefficients of $C(x)^2$ is comparable in size — see if you can figure out why.)
