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 $$a_n^2<\alpha^2+2n+\frac{1}{\alpha^2}+\frac{1}{2}ln\left ( \frac{2n}{\alpha^2}+1 \right )$$
?
I'll really appreciate your help. Thanks.
 A: 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. 
A: This is something I typed on October 31. It goes with an old question I can't find. It does indicate that both lower and upper bounds for $a_n$ can probably be proven by induction, and should be considered simultaneously. I will think about it.
An approach:
Suppose we have continuous $f(x) $ for $x > -1,$ with $f > 0,$ $f' < 0,$ and $$\lim_{x \rightarrow +\infty} f(x) = 0.$$
Take a sequence with, say, $x_0 = 0,$ after which $x_{n+1} = x_n + f(x_n).$ Then $x_n$ grows without bound. 
This is a little counterintuitive. It is an exercise in some famous book. 
I think the desired answer is this: it is an increasing sequence, assuming it has an upper bound, it has a least upper bound $L,$ which is also its limit. 
However, for some large $n,$ we have 
$x_n > L - \frac{f(L)}{2},  $ so that $x_{n+1} > L + \frac{f(L)}{2},  $ contradicting $L$ being an upper bound and the assumption that there is any upper bound.
My Proof: however small it might be, $f(1) > 0.$ As long as $x_n < 1,$ we get $x_{n+1} > x_n + f(1).$ Therefore, if we take 
$$ N_1 = \left\lceil \frac{1}{f(1)} \right\rceil,  $$ we have 
$x_n \geq 1$ for $n \geq N_1.$
However small it might be, $f(2) > 0.$ As long as $x_n < 2,$ we get $x_{n+1} > x_n + f(2).$ Therefore, if we take 
$$ N_2 = \left\lceil \frac{1}{f(2)} \right\rceil,  $$ we have 
$x_n \geq 2$ for $n \geq N_1 + N_2.$
Next $x_n \geq 3$ for $n \geq N_1 + N_2 + N_3.$
And so on, forever. 
$$  \bigcirc    \bigcirc   \bigcirc   \bigcirc   \bigcirc   \bigcirc   \bigcirc  $$
In comparison, what about a sequence that really does converge, where we merely say 
$$ y_{n+1} > y_n. $$
For example, just take
$$ y_n = 1 - \frac{1}{n}. $$ This has limit $1.$ We really could work out some continuous function $g$ such that 
$$g(y_n) = y_{n+1} - y_n = \frac{1}{n} - \frac{1}{n+1} =  \frac{1}{n^2 + n}.$$ Which seems interesting, until we point out that 
continuity of $g$ demands $$ g(1) = 0. $$ That is, the strict positivity of $f$ is crucial in the result proved above.
