For each generating function $A(x)$, find c and ρ such that the coefficient $a_n$ of $x^n$ in the generating function $A(x)$ satisfy: $a_n \sim c * ρ^n$.
$$A(x)= \frac{1-2x}{(1-x)(1-3x)}$$
$$A(x)=\frac{1}{\prod_{i=1}^{k} (1-ix)}$$
$$A(x)=\frac{e^x}{(1-2x)^2}$$
$$A(x)=\frac{3}{4-e^x}$$
My work:
First g.f
We have two poles 1 and 1/3. The smallest pole is 1/3. So ρ=3 and to find c, we multiple A(x) by (1-3x) and take a limit of (1-3x)A(x) where $x\to 1/3$, getting c= 1/3 / 2/3 = 1/2 Thus, $a_n \sim 1/2*3^n$
Second g.f
We can write it as $A(x)=1/(1-x)(1-2x)....(1-kx)$. The poles are 1,1/2,...1/k , we have k simple poles and the smallest pole is 1/k so ρ=k , then multiple A(x) by (1-kx) and take a limit of $(1-kx)A(x)$ where $x\to 1/k$ getting $c=k^{k-1} / (k-1)!$ therefore, a_n~c * ρ^n for c and ρ which I calculated.
Third g.f
There is one pole 1/2 Therefore ρ=2 ,We can write A(x) as: $A(x)=e^x \sum_{n\geq 0} 2^nx^n {n+1 choose n}$. Then substitute x=1/2 in the analytic exponential function e^x. Get $A(x)= e^{1/2} \sum_{n\geq 0} 2^n x^n {{n+1} \choose {n}}$. Thus $A(x)$ behaves as $e^{1/2} 2^n {{n+1} \choose {n}}=e^{1/2}* 2^n *(n+1) \sim e^{1/2} * 2^n * n$ so in this case $a_n \sim c* ρ^n * n^{\alpha}$.Where $\alpha=1$.
forth g.f
the only pole which is simple is $\log 4$ thus $ ρ=1/ \log 4 $. And $A(x)=3(x-\log 4)/ (4-e^x)(x- \log 4)$ . So if we denote $A(x)=\sum_{n\geq 0} a'_n x^n/n!$ we get, $a_n=a'_n/n! \sim 3/4 * 1/(\log 4)^{n+1}$.
Can you please evaluate my method, provide any improvements or correct me if there is need to.
