does this sum have a limit?

For each specific $k$ we have $$\binom{2k}k\binom{n+1}k\frac{k-1}{2^k\binom{4n}k}\to \binom{2k}k\frac{k-1}{8^k},$$ that suggests that the sum tends to $$\sum_{k=2}^\infty \binom{2k}k\frac{k-1}{8^k}=1-\frac{1}{\sqrt 2},$$ as $$\sum_{k= 0}^\infty \binom{2k}k x^k =(1-4x)^{-1/2};\,\\\sum_{k= 0}^\infty k\binom{2k}k x^k =(x\frac{d}{dx})(1-4x)^{-1/2}=2x(1-4x)^{-3/2};\\ \sum_{k= 0}^\infty (k-1)\binom{2k}k x^k =(6x-1)(1-4x)^{-3/2}=|_{x=1/8}-\frac1{\sqrt{2}}.$$

In order to justify this limit changes (limit of a sum is a sum of limits) we majorate each term in our sum. Say, $$\frac{\binom{n+1}k}{\binom{4n}k}=\frac{(n+1)n(n-1)\dots(n-k+2)}{4n(4n-1)(4n-2)\dots (4n-k+1)}<4^{-k}(1+\frac{1}n)(1+\frac{1}{4n-1})<3\cdot 4^{-k}.$$ this allows to estimate the total sum of summands with $k\geqslant k_0$ by $3\sum_{k\geqslant k_0} \binom{2k}k\frac{k-1}{8^k}$, that tends to 0 for large $k_0$. Therefore the limit of your sum is indeed a sum of limits, moreover, the estimates are very specific and should prove that $0<a_n<1$ also after checking some small values of $n$.

Not an answer. First, Mathematica says that $$a_n = \frac{-4 n \, _2F_1\left(\frac{1}{2},-n-1;-4 n;2\right)+n \, _2F_1\left(\frac{3}{2},-n;1-4 n;2\right)+\, _2F_1\left(\frac{3}{2},-n;1-4 n;2\right)}{4 n}.$$

Secondly, for all reasonable values of $n$ it seems that $a_n < -0.7,$ and, indeed, I conjecture that $$\lim_{n\rightarrow \infty} a_n = -\frac{1}{\sqrt{2}}.$$

Now, this is a response to the original form of this question, with the sum going from $0.$ The first two terms contribute $-1$ to the sum, so for the current version, the conjectured limit is $$1-\frac{1}{\sqrt{2}}.$$