Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence $(a_n)$ is convergent.
This is what I did:
We have:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
$$a_n = n\sqrt{n^2+n}-an^2+n\sqrt{n^2-n}$$
$$a_n = n^2\sqrt{1+\dfrac{1}{n}} + n^2 \sqrt{1 - \dfrac{1}{n}} - an^2$$
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$
The only way we could have $a_n$ convergent is if we would have the limit result in an indeterminate form. In this case, we need $\infty \cdot 0$. So we have:
$$\lim_{n \to \infty} \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg ) = 0$$
And from that we can conclude that:
$$a = 2$$
So now that I found $a$, I must find the limit of the sequence. So this limit:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
I tried this:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n^2+n}-2n^2+n\sqrt{n^2-n}$$
$$= \lim_{n \to \infty}( n\sqrt{n^2+n} - n^2) + \lim_{n \to \infty} (n\sqrt{n^2-n} -n^2)$$
And then I multiplied both of those limits with its respective conjugate, but after my calculations, it still results in an indeterminate form, only it's $\infty - \infty$ this time.
So, if my previous calculations aren't wrong, my question is how can I find this limit:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
 A: Your initial working leads well into using the generalized binomial expansion which gives the asymptotic behaviour as $n\to\infty$
\begin{align}
a_n
&=n^2\left(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}-2\right)\\
&\sim n^2\left(1+\frac1{2n}-\frac1{8n^2}+\dots+1-\frac1{2n}-\frac1{8n^2}-\dots-2\right)\\
&=-\frac14\\
\end{align}
Further terms of this expansion gives the more precise asymptotic behaviour of
\begin{align}
a_n
&\sim-\frac14-\frac5{64n^2}-\frac{21}{512n^4}-\dots\\
&=-\sum_{k=0}^\infty\frac{\binom{4k+1}{2k}}{4^{2k+1}(k+1)n^{2k}}\qquad(n\ge1)\\
\end{align}
A: Hint
Rationalizing you get
$$\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1}= (\sqrt{n + 1} - \sqrt{n})-(\sqrt{n} - \sqrt{n - 1})\\=\frac{1}{\sqrt{n+1}+\sqrt{n}}- \frac{1}{\sqrt{n}+\sqrt{n-1}}\\
=\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n}+\sqrt{n-1})}$$
Repeat:
$$\sqrt{n-1}-\sqrt{n+1}=\frac{-2}{\sqrt{n-1}+\sqrt{n+1}}$$
A: $\begin{array}\\
\Delta^2_hf(x)
&= f(x-h)-2f(x)+f(x+h)\\
\text{so}\\
\Delta^2_1f(x)
&= f(x-1)-2f(x)+f(x+1)\\
&\approx  (f(x)-f'(x)+f''(x)/2-f'''(x)/6+...)-2f(x)+(f(x)+f'(x)+f''(x)/2+f'''(x)/6+...)\\
&\approx  f''(x)/2+O(f''''(x))\\
\end{array}
$
If 
$f(x) = x^{1/2},\\
f'(x) = x^{-1/2}/2,\\
f''(x) = -x^{-3/2}/4,\\
f'''(x) = 3x^{-5/2}/8,\\
f''''(x) = -15x^{-7/2}/16,\\
$
so
$\Delta^2_1(\sqrt{n})
=-n^{-3/2}/4+O(n^{-7/2})
$
so
$n^{3/2}\Delta^2_1(\sqrt{n})
=-1/4+O(n^{-2})
$.
A: You could have done everything a bit faster. Starting from what you wrote
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$ let $x=\frac 1 n$, use the binomial expansion or Taylor series, replace $x$ by $\frac 1n$ to get
$$a_n=(2-a) n^2-\frac{1}{4}-\frac{5}{64 n^2}+O\left(\frac{1}{n^4}\right)$$ which gives the results $a=2$, the limit and how the limit is approached.
For your curiosity, using @Peter Foreman's result for $a=2$
$$a_n =-\sum_{k=0}^\infty\frac{\binom{4k+1}{2k}}{4^{2k+1}(k+1)n^{2k}}=\left( \sqrt{2\sqrt{1-\frac{1}{n^2}}+2}-2\right) n^2$$
