Convergence of sequence $\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-a\sqrt{n}+\sqrt{n-1})$ I am able to show that the sequence converge only if $a=2$, I also did a numeric analysis of the sequence, and I am sure that the value of this limit is $-\frac{1}{4}$.  However, I do not know how to prove that
$\lim_{n \rightarrow  \inf }n\sqrt{n}(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})=-\frac{1}{4}$

For an analytic approach, we can use a higher order version of MVT.
When $a = 2$, the expression
$$\sqrt{n-1} - 2\sqrt{n} + \sqrt{n+1} = \left. \sum_{k=0}^{2}(-1)^k\binom{2}{k} \sqrt{x+k}\;\right|_{x = n-1}$$
is the second order finite differences of the function $x \mapsto \sqrt{x}$ at $n-1$. 
The MVT we need has the form:

For any $f(x)$ defined on $[a,a+mh]$ continuously differentiable up to $m$ times, there exists a $b \in (a,a+mh)$ such that:
  $$\left.\frac{d^m f(x)}{dx^m}\right|_{x=b}
= \left.\frac{1}{h!}\Delta^n_h[f](x)\right|_{x=a}
\stackrel{def}{=} \frac{1}{h!} \sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}f(a+kh)
$$

For a proof of above, see this answer.
Apply this MVT to the function $\sqrt{x}$ at $a = n-1$ for any $n > 1$, with $h = 1$ and $m = 2$.  
Notice $\sqrt{x}'' = -\frac14 x^{-3/2}$, we find:
$$n^{3/2}\left(\sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right)
= -\frac14\left(\frac{n}{b_n}\right)^{3/2}
\quad\text{ for some }\quad b_n \in (n-1,n+1)
$$
Since $\left(\frac{n}{n+1}\right)^{3/2} < \left(\frac{n}{b_n}\right)^{3/2} < 
\left(\frac{n}{n-1}\right)^{3/2}$ and terms on both sides converge to $1$ as $n \to \infty$,
we can conclude
$$\lim_{n\to\infty}n^{3/2}\left(\sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right) = -\frac14$$
