Prove there exists $\lim_\limits{n\to \infty}{a_n}$ and find $\lim_\limits{n\to \infty}{a_n\over n}$. Let $(a_n)$ be defined by $a_1=1$, $a_{n+1}=a_n+{1\over a_n}$. Prove there exists $\lim_\limits{n\to \infty}{a_n}$ and find $\lim_\limits{n\to \infty}{a_n\over n}$. 
I tried many things but they wouldn't work. I would appreciate your help.
 A: Just for a reference, we can prove that
$$ a_n = \sqrt{2n} + \frac{\log n}{4\sqrt{2n}} + O \left( \frac{1}{\sqrt{n}} \right). $$
The idea is essentially the same as @LeGrandDODOM's proof, but in this case we successively improve the error term.
A: Actually, the limit $\lim a_n$ does not exist. For, assume it does though, and let it equal $\ell$. Then, $a_{n + 1} = a_n + 1/a_n$ yields $\ell = \ell + 1 / \ell$, a contradiction.
We now show that $\lim a_n/n$ exists. Since for every $n$ we know that $0 < a_n < n$, we get that $0 < a_n/n < 1$. Observe also that $a_n/n$ is a decreasing sequence. Thus, $a_n/n$ is bounded and monotone, so it has a limit.
Also we can prove that for any $N$ there exists $\varepsilon_N > 0$ such that for all $n > N$ we have $a_n < n \varepsilon_N$. This would prove that $a_n/n \to 0$.
A: *

*By induction, prove $a_n \geq 1$ for all $n$.

*Use this to show $(a_n)_n)$ is strictly increasing.

*By monotonicity, it must converge in $\mathbb{R}\cup\{+\infty\}$.

*By continuity of the function $f\colon x > 0 \mapsto x+\frac{1}{x}$, if the limit $\ell$ exists it must satisfy $\ell = f(\ell)$. What is the only possible limit?


Then, once you known $a_n\xrightarrow[n\to\infty]{} \infty$, you can consider 
$$\ln a_n = \ln a_n - \ln a_1  = \sum_{k=1}^{n-1} \ln\frac{a_{k+1}}{a_k} = \sum_{k=1}^{n-1} \ln\!\left(1+\frac{1}{a_k^2}\right)$$.
You know that the LHS goes to $\infty$, so the RHS must be a divergent sequence (to infinity). Since $a_n\to \infty$, $\ln\!\left(1+\frac{1}{a_k^2}\right)\sim_{n\to\infty} \frac{1}{a_k^2}$ and by comparison of divergent series this implies that the series
$$
\sum \frac{1}{a_k^2}
$$
diverges as well. Why do that imply that the positive sequence $\frac{a_n}{n}$ cannot have any non-zero limit? And why does that imply that then it must have limit $0$?
A: As the second part of your question suggests, and confirmed by one of the comments, the limit must be $\infty$ because your sequence is strictly increasing and a limit $L$ would have to satisfy $L = L + 1/L$ which is impossible. 
For the second part of the question, note that your sequence grows on the order of $\log n$ or smaller (this is fairly easy to show if you know that the harmonic series grows like $\log n$). Thus the limit of $a_n/n$ is $0$.
A: As suggested in the comments, $\lim_{n\to\infty}a_n= \infty$.
Note that $a_{n+1}^2=a_n^2+2+\frac{1}{a_n^2}$
Hence $\lim_{n\to\infty} (a_{n+1}^2-a_n^2)=2$
By Cesaro theorem, $\displaystyle \lim_{n\to\infty}\frac{\sum_{k=1}^n a_{k+1}^2-a_{k}^2}{n}=2$, that is to say $\lim_{n\to\infty}\frac{a_n^2}{n}=2$
Hence $\lim_{n\to\infty} \frac{a_n}{n}=0$
