<s>[INST] Upper bound of $a_{n+1}=a_n + \frac{1}{a_n}$ How can I prove:
If $$a_0=\alpha>0\quad and\quad a_{n+1}=a_n + \frac{1}{a_n}$$, then $$a_n^2<\alpha^2+2n+\frac{1}{\alpha^2}+\frac{1}{2}ln\left ( \frac{2n}{\alpha^2}+1 \right )$$
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I'll really appreciate your help. Thanks.
 [/INST] We have $\displaystyle a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$. Hence for $n\geq 1$
$$a_n^2=a_0^2+2n+\sum_{k=0}^{n-1}\frac{1}{a_k^2}$$
This imply that $\displaystyle a_k^2\geq a_0^2+2k$ for $k\geq 1$. Hence
$$a_n^2\leq a_0^2+2n+\frac{1}{a_0^2}+\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}$$
Now $$\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}\leq \int_0^{n}\frac{dt}{a_0^2+2t}=\frac{1}{2}\log (\frac{2}{a_0^2}n+1)$$
 and we are done. 
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