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
$\lim_{n \rightarrow  \inf }n\sqrt{n}(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})=-\frac{1}{4}$

You can consider $n=1/x$ and study the limit of the function at $0$ from the right:
$$
\lim_{x\to0^+}
\frac{1}{x}\frac{1}{\sqrt{x}}
\left(\sqrt{\frac{1}{x}+1}-a\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{x}-1}\right)
=
\lim_{x\to0^+}\frac{\sqrt{1+x}-a+\sqrt{1-x}}{x^2}
$$
If $a\ne2$, the limit is infinite, because the limit of the numerator is $2-a$ (and this decides between $\infty$ and $-\infty$).
Thus we can assume $a=2$.
Now you can rationalize; first multiply by $\sqrt{1+x}-2-\sqrt{1-x}$:
$$
\lim_{x\to0^+}\frac{(1+x+4-4\sqrt{1+x})-(1-x)}{x^2(\sqrt{1+x}-2-\sqrt{1-x})}
$$
Isolating the part that has limit $-2$ and simplifying the $-2$ in the numerator, we reduce to
$$
\lim_{x\to0^+}\frac{2\sqrt{1+x}-x-2}{x^2}
\cdot
\lim_{x\to0^+}\frac{-2}{\sqrt{1+x}-2-\sqrt{1-x}}
$$
The second factor has limit $1$ and can be disregarded. Now you can rationalize again:
$$
\lim_{x\to0^+}\frac{2\sqrt{1+x}-x-2}{x^2}=
\lim_{x\to0^+}\frac{4(1+x)-(x^2+4x+4)}{x^2(2\sqrt{1+x}+x+2)}
$$
and finish up.
With a Taylor expansion, recalling that
$$
\sqrt{1+t}=1+\frac{1}{2}t-\frac{1}{8}t^2+o(t^2)
$$
we have
$$
\lim_{x\to0^+}\frac{\sqrt{1+x}-2+\sqrt{1-x}}{x^2}=
\lim_{x\to0^+}\frac{(1+x/2-x^2/8)-2+(1-x/2-x^2/8)+o(x^2)}{x^2}=-\frac{1}{4}
$$
