Let sequence $\{a_{n}\}$ such $$\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}},n\ge 1$$ Find limits $$\lim_{n\to\infty}n\left(\dfrac{9}{16}-a_{n}\right)$$
I am working on a problem and I am lead to prove the following inequality $$\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}\ge 2\sqrt{n}$$ and $$\sqrt{na_{n}+n}+\sqrt{na_{n}}\le 2\sqrt{n}$$
We have $\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}}$ so
$(\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1})\dfrac{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}\le\dfrac{\sqrt{n}}{2}\le(\sqrt{na_{n}+n}-\sqrt{na_{n}})\dfrac{\sqrt{na_{n}+n}+\sqrt{na_{n}}}{\sqrt{na_{n}+n}+\sqrt{na_{n}}}$
giving $\dfrac{n}{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}\le\dfrac{\sqrt{n}}{2}\le\dfrac{n}{\sqrt{na_{n}+n}+\sqrt{na_{n}}}$.
Then $\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}\ge 2\sqrt n\ge\sqrt{na_n+n}+\sqrt{na_n}$.
It comes $na_n+n+1\ge 4n+na_n+1-4\sqrt n\sqrt{na_{n}+1}$ or else $4\sqrt n\sqrt{na_{n}+1}\ge 3n$ and $4n-4\sqrt n\sqrt{na_n}+na_n\ge na_n+n$ or else $3n\ge 4\sqrt n\sqrt{na_n}$.
Thus $a_n\leq\dfrac{9}{16}\leq a_n+\frac{1}{n}$.
Now if $a_n=\dfrac{9}{16}$, the sought limit is 0 but if $a_n=\frac{9}{16}-\frac{1}{n}$, the limit is 1.
