I was having trouble evaluating the follwing limit: $\lim\_{n\to\infty}\frac{4^n-C(2n+1,n)}{2^n}.$ By simply plugging in values I know the limit should equal infinity but my showing this rigorously was difficult. First I expanded out and got,
\begin{align\*}
\lim\_{n\to\infty}\frac{4^n-\frac{(2n+1)!}{n!(n+1)!}}{2^n}.
\end{align\*}
Next I tried approached it two ways:

First I tried taking the limit of the difference portion which is:
\begin{align\*}
\lim\_{n\to\infty} \frac{4^n}{2^n}-\lim\_{n\to\infty}\frac{\frac{(2n+1)!}{n!(n+1)!}}{2^n}
\end{align\*}
but this was trouble some since I ended up with an $\infty-\infty$ situation.

Secondly, I tried using L'Hôpital's rule which gives:
\begin{align\*}
\lim\_{n\to\infty} \frac{\frac{d}{dn}\bigg(4^n-\frac{(2n+1)!}{n!(n+1)!}\bigg)}{\frac{d}{dn}2^n}.
\end{align\*}
This turned out to be trouble some since I ended up with 
\begin{align\*}
\lim\_{n\to\infty} \frac{4^nlog(4)-\frac{d}{dn}\big(\frac{(2n+1)!}{n!(n+1)!}\big)}{2^nlog(2)}.
\end{align\*}
solving for $\frac{d}{dn}\big(\frac{(2n+1)!}{n!(n+1)!}\big)$ lead me down the rabbit hole of gamma functions and digamma functions which lead no where for me.

Any advice on how I should approach evaluating the following limit? I feel as though I am missing something and I am not clever enough to continue down the rabbit hole of the second approach to find any results. Thanks!