<s>[INST] 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:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
 [/INST] 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:
$$\sqrt{n-1}-\sqrt{n+1}=\frac{-2}{\sqrt{n-1}+\sqrt{n+1}}$$
</s>