Define the sequence given by the finite sum
$$a\_n:=\sum\_{k=2}^{n+1}\binom{2k}k\binom{n+1}k\frac{k-1}{2^k\binom{4n}k}.$$



> 
> **Questions.**
> 
> 
> (1) Is $0<a\_n<1$?
> 
> 
> (2) Does the limit $\lim\_{n\rightarrow\infty}a\_n$
> exist? If yes, what is its value?
> 
> 
> 



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 $n$.


