$f_n(x)=(\sum_{p=1} ^{n} \frac{1}{\sqrt{nx+p}} )-\sqrt{n}$ let $f_n(x)=(\sum_{p=1} ^{n} \frac{1}{\sqrt{nx+p}} )-\sqrt{n}$
i had proven that that the root  of $f_n(x)$ is $a_n$ unique and on the interval $[0,1]$
the question is how to show that $ \lim_{x \to +\infty} a_n = \frac{9}{16} $
please help me with this question
 A: Note for any $x$ and any $n\in\Bbb N$ that: $$f_n(x)=\sqrt{n}\left(-1+\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{x+\frac{k}{n}}}\right)$$So, in particular: $$1=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{a_n+\frac{k}{n}}}$$You can use this to derive bounds $0\le a_n\le1$ for all $n\in\Bbb N$.

If $a_n>1$, for some $n$, then we would have: $$1<\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{1}}=1$$Which is false. If $a_n<0$ for some $n$ then we would have: $$1>\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{k/n}},\,\sqrt{n}>\sum_{k=1}^n\frac{1}{\sqrt{k}}$$This last inequality is evidently false for $n=1$. If it is false for all $1\le n'\le n-1$, but true for $n$, that would mean: $$\sqrt{n}-\frac{1}{\sqrt{n}}>\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}>\sqrt{n-1}$$Or equivalently: $$(n-1)^2>n(n-1),\,n-1>n$$Which is also false. So, by induction, $a_n<0$ can never occur.

This reveals the recurrence: $$1=\frac{1}{\sqrt{a_{n+1}+1}}+\sum_{k=1}^n\left(\frac{1}{\sqrt{a_{n+1}+\frac{k}{n+1}}}-\frac{1}{\sqrt{a_n+\frac{k}{n}}}\right)$$Which is valid for all $n\in\Bbb N$ (by multiplying by $n$ and subtracting) and this shows that, if $a_{n+1}\le a_n$ for some $n$ we'd have the right sum strictly positive and thus the RHS as strictly greater than $1$, a contradiction; $a_n<a_{n+1}$ follows for all $n\in\Bbb N$. By the monotone convergence theorem, the $a_\bullet$ have a limit $\alpha\in[0,1]$.
We then know by this convergence and by the Riemann sum that: $$1=\lim_{n\to\infty}1=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{a_n+\frac{k}{n}}}=\int_\alpha^{\alpha+1}\frac{1}{\sqrt{x}}\,\mathrm{d}x$$
To justify this rigorously, note that: $$\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{\alpha+\frac{k}{n}}}\to\int_{\alpha}^{\alpha+1}\frac{1}{\sqrt{x}}\,\mathrm{d}x$$And: $$\frac{1}{n}\sum_{k=1}^n\left(\frac{1}{\sqrt{\alpha+\frac{k}{n}}}-\frac{1}{\sqrt{a_n+\frac{k}{n}}}\right)\to0$$By the dominated convergence theorem applied to the counting measure space.
So we have: $$\frac{1}{2}=\sqrt{\alpha+1}-\sqrt{\alpha}$$I don't think it's obvious by inspection, but this has a unique solution (uniqueness stemming from the fact that $x\mapsto\sqrt{x+1}-\sqrt{x}$ is a decreasing function $(0,\infty)\to(0,\infty)$ - derivative test) of $\alpha=\frac{9}{16}$.
You can actually explicitly solve the last equation - by difference of two squares we get: $$\sqrt{\alpha+1}+\sqrt{\alpha}=2$$Then we take the difference to find: $$\frac{3}{4}=\frac{1}{2}\cdot\left(2-\frac{1}{2}\right)=\sqrt{\alpha}\implies\alpha=\frac{9}{16}$$
A: I tried to simplfy the part of the solution related to the Riemann sum... (I did! It is really a Rieamann sum trick.)
$f_n(x)=0\implies \sum_{p=1}^{n}\frac{1}{\sqrt{nx+p}}=\sqrt{n}\implies \sum_{p=1}^{n}\frac{1}{n}\frac{1}{\sqrt{x+\frac{p}{n}}}=1\implies$
$\int_0^1\frac{dy}{\sqrt{x+y}}=1\implies 2\sqrt{x+1}-2\sqrt{x}=1\implies x=\frac{9}{16}.$
