A second order recurrence relation problem I was asked by a friend with this problem but I can't solve it. Can anyone help?
We have a sequence $\left\{a_n\right\}$ that satisfies $a_1=1$, $a_2=2$,
$$a_n+\frac{1}{a_n} =\frac{a_{n+1}^2+1}{a_{n+2}}$$
where $n$ is a positive integer.
Prove that 


*

*$a_{n+1}=a_n+\frac{1}{a_n}$

*$2n-1\le a_n^2\le3n-2$

*Let $S_n$ be the sum of the sequence $\left\{1\over a_n\right\}$, prove $62<S_{2014}<77$.


(a)
By induction, assume $a_{k+1} =  a_k + \frac1{a_k}$. Then to get the result for $n=k+1$, consider
$$\begin{align*}
a_k+\frac1{a_k} =& \frac{a_{k+1}^2+1}{a_{k+2}}\\
a_{k+1}=&\frac{a_{k+1}^2+1}{a_{k+2}}\\
a_{k+2} =& a_{k+1}+\frac1{a_{k+1}}
\end{align*}$$
Also prove the base case for $n=1$ holds.
(b)
Using $a_{n+1}^2 = a_{n}^2+2+\dfrac1{a_n^2}$ and by induction, assume $2k-1\le a_k^2 \le 3k-2$. Then for $n=k+1$, the first inequality is
$$a_{k+1}^2 = a_k^2 + 2 +\frac1{a_k^2}\ge2k-1+2 = 2(k+1)-1$$
and the second inequality is similar. Use the fact that $0\le\dfrac1{a_k^2}\le1$.
(c)
$$\begin{array}{rcl}
2n+1\le&a_{n+1}^2&\le3n+1\\
\sqrt{2n+1}\le& a_{n+1} &\le \sqrt{3n+1}\\
\sqrt{2n+1}\le& a_1+\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}&\le \sqrt{3n+1}\\
\sqrt{2\times2014+1}-a_1 \le& \sum_{i=1}^{2014}\frac1{a_i} &\le \sqrt{3\times2014+1}-a_1\\
62.4\cdots \le& \sum_{i=1}^{2014}\frac1{a_i} &\le 76.7\ldots\\
\end{array}$$
