Haw to find the asymptotic behavior of partial sum of series? What is the asymptotic behavior of the partial sum of Taylor series?
$$[x^n]f(x)(g(x))^k$$
where $f(x)=\frac{1}{1-4x}$ and $g(x)=\frac{1-\sqrt{1-4x}}{2x}$.
Notice that:
$$(I)\quad f(x)=\sum\limits_{s=0}^\infty4^sx^s$$
$$(II)\quad g(x)=\sum\limits_{s=0}^\infty \frac{1}{s+1}\binom{2s}{s}x^s$$
$$(III)\quad (g(x))^k=\sum\limits_{s=0}^\infty \frac{k}{2s+k}\binom{2s+k}{s}x^s$$
In other words, while $k=n$, my question is what is the asymptotic behavior of
$$[x^n]f(x)(g(x))^n=\sum\limits_{s=0}^n 4^{n-s} \frac{n}{2s+n}\binom{2s+n}{s}\sim ?$$
Or more generally, when $k=\lfloor n^\alpha\rfloor ,0\leq\alpha\leq 1$.
$$[x^n]f(x)(g(x))^k=\sim ?$$
Any help would be appreciated!
 A: This is just a sketch, and I haven't checked all of the details.
Let
$$
S_n := \sum_{s=0}^n 4^{n-s} \frac{n}{2s+n}\binom{2s+n}{s}.
$$
The main difficulty is the binomial coefficient. It seems like $S_n$ is dominated by the terms where $s \approx n$, so we expect that we should have
$$
S_n \approx \sum_{s=0}^n 4^{n-s} \frac{n}{2s+n}\binom{3n}{n}
$$
in some sense, but if we compare plots of these they seem a bit off.
By carefully estimating the binomial coefficient we can prove that
$$
\binom{2s+n}{s} \sim \binom{3n}{n} \left(\frac{2}{9}\right)^{n-s} \exp\!\left\{-\frac{(n-s)^2}{12n}\right\} \tag{$*$}
$$
when $s = n - o(n^{2/3})$. This interval for $s$ should be plenty wide enough for our purposes. Thus we get
$$
\begin{align}
S_n &\sim \binom{3n}{n} \sum_{s=0}^{n} \frac{n}{n+2s} \left(\frac{8}{9}\right)^{n-s} \exp\!\left\{-\frac{(n-s)^2}{12n}\right\} \\
&\sim \binom{3n}{n} \sum_{s=0}^{n} \frac{n}{n+2n} \left(\frac{8}{9}\right)^{n-s} \exp\!\left\{-\frac{(n-n)^2}{12n}\right\} \\
&\sim \frac{1}{3} \binom{3n}{n} \sum_{r=0}^{\infty} \left(\frac{8}{9}\right)^r \\
&= \,3\,\binom{3n}{n},
\end{align}
$$
which agrees with numerical experiments.
Generalizing this to other $k$ would require other estimates like $(*)$. Please let me know if you would like me to include the details of that asymptotic.
