<s>[INST] How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$ 
let sequence $\{a_{n}\}$ such $a_{1}=1$,and 
  $$a_{n+1}a_{n}=n,n\ge 1$$
  show that
  $$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$

(2): I consider we can find this limit
$$\lim_{n\to+\infty}\dfrac{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}{\sqrt{n}}=?$$
My try:since
$$a_{n+2}a_{n+1}-a_{n+1}a_{n}=n+1-n=1$$
so
$$a_{n+2}=\dfrac{1}{a_{n+1}}+a_{n}$$
so
$$\dfrac{1}{a_{n+1}}=a_{n+2}-a_{n}$$
so
\begin{align*}\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}&=a_{1}+(a_{3}-a_{1})+(a_{4}-a_{2})+(a_{5}-a_{3})+\cdots+(a_{n+1}-a_{n-1})\\
&=a_{1}+a_{n+1}+a_{n}-a_{1}-a_{2}\\
&=a_{n+1}+a_{n}-a_{2}
\end{align*}
since $$a_{1}=1,a_{1}a_{2}=1\Longrightarrow a_{2}=1$$
so
$$a_{n+1}+a_{n}-a_{2}\ge2\sqrt{a_{n+1}a_{n}}-1\ge 2\sqrt{n}-1$$
so left hand inequality is prove it.Now consider Right hand inequality,we only prove this
$$a_{n}+a_{n+1}<\dfrac{5}{2}\sqrt{n}$$
since
$$a_{n}a_{n+1}=n\Longrightarrow a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$$
so maybe we can $a_{n}<2\sqrt{n}$?
becasue $$\dfrac{5}{2}\sqrt{n}=2\sqrt{n}+\dfrac{n}{2\sqrt{n}}$$
so let $$f(x)=x+\dfrac{n}{x}$$
if we can prove $a_{n}<2\sqrt{n}$,Then
$$f(x)\le f(2\sqrt{n})=\dfrac{5}{2}\sqrt{n}$$
Thank you very much 
 [/INST] observe that for $a_{n}$ to satisfy the inequality $a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$
 we need      $\dfrac{\sqrt{n}}{2} <a_{n}<2\sqrt{n}$.
this can be proved by induction: first observe that it is true for $a_{1}$ and $a_{2}$ then suppose it is true for any n ($a_{n}$) then prove it is true for n+1 ($a_{n+1}$) and you are done 
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