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
$\lim_{n \rightarrow  \inf }n\sqrt{n}(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})=-\frac{1}{4}$
 A: $$\sqrt{n\pm1}=\sqrt n\left(1\pm\frac1n\right)^{1/2}
=\sqrt n\left(1\pm\frac1{2n}-\frac{1}{8n^2}+O(n^{-3})\right)$$
(binomial theorem).
So
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt n=\sqrt n\left(-\frac1{4n^2}+O(n^{-3})\right)$$
etc.
A: 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
$$\lim_{n\to\infty}n^{3/2}\left(\sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1}\right) = -\frac14$$
A: The best option to solve this is based on limited expansions, following the indications in @LordShark's answer. For an almost purely algebraic approach, note that the $n$th term $$x_n=n\sqrt{n}\left(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\right)$$ is also equal to $$x_n=\frac{-2n\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n-1})}$$ hence $$x_n=\frac{-2}{\left(\sqrt{1+\frac1n}+1\right)\left(1+\sqrt{1-\frac1n}\right)\left(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}\right)}$$ whose limit should be clear.
The alternative formula for $x_n$ given above is based on the identities $$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}$$ and $$\sqrt{n}-\sqrt{n-1}=\frac1{\sqrt{n}+\sqrt{n-1}}$$
hence the "analytical" part of this approach is reduced to the fact that $$\lim_{x\to0}\sqrt{1+x}=1$$ that is, to the continuity of the square root function at $1$.
A: 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
$$
\lim_{x\to0^+}\frac{\sqrt{1+x}-2+\sqrt{1-x}}{x^2}=
\lim_{x\to0^+}\frac{(1+x/2-x^2/8)-2+(1-x/2-x^2/8)+o(x^2)}{x^2}=-\frac{1}{4}
$$
