Compute limit of a logarithmic term Let $k\in\mathbb{N}$.
Is it possible to determine
$$
\lim_{n\to\infty}\frac{1}{n}\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right)?
$$
Can write the log-term as sum,
\begin{align*}
&\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right)\\
&=\sum_{j=k}^{n+k}\log(4)+\sum_{j=k}^{n+k}\log(j)+\log(k-1)+\sum_{i\in 2\mathbb{N}_0+1}^{2n-3}\log(k+i)
\end{align*}

*

*The middle term tends to zero when dividing by $n$ and letting $n\to\infty$, i.e.,
$\frac{1}{n}\log(k-1)\to 0$.


*The first sum is $\sum_{j=k}^{n+k}\log(4)=n\log(4)$, so that $\frac{1}{n}\sum_{j=k}^{n+k}\log(4)=\log(4)$.
 A: We have for $k\in\Bbb N\setminus\{1\}$ (since $k=1$ results in a zero inside the logarithm):
$$
L=\lim_{n\to\infty}\frac{1}{n}\log\left(\left(\prod_{j=0}^n 4(k+j)\right)(k-1)(k+1)(k+3)\cdots(k+2n-3)\right),
$$
which using the Pochhammer symbol can be simplified as
$$
\begin{aligned}
L
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^{n+1}(k+1)_n(k)_{2n-2}k(k-1)\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^n\Gamma(n+k+1)\Gamma(2n+k-2)\frac{4k(k-1)}{k!\Gamma(k)}\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^n\Gamma(n+k+1)\Gamma(2n+k-2)\right).
\end{aligned}
$$
Now using asymptotic expansions for the gamma function we further write
$$
\begin{aligned}
L
&=\lim_{n\to\infty}\frac{1}{n}\log\left(4^ne^{-3n}n^{n+k+1/2}(2n)^{2n+k-5/2}(2\pi)\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\left(n\log4-3n+(n+k+1/2)\log n+(2n+k-5/2)\log(2n)\right)\\
&=\lim_{n\to\infty}\left(\log 4-3+\frac{n+k+1/2}{n}\log n+\frac{2n+k-5/2}{n}\log(2n)\right)\\
&=\log 4-3+\lim_{n\to\infty}\left(\log n+2\log(2n)+\underbrace{\frac{k+1/2}{n}\log n+\frac{k-5/2}{n}\log(2n)}_{\to 0}\right)\\
&=\log 4-3+\lim_{n\to\infty}\left(\log n+2\log(2n)\right)\\
&=\infty.
\end{aligned}
$$
