Does the recursive sequence $a_1 = 1, a_n = a_{n-1}+\frac{1}{a_{n-1}}$ converge? Does the recursive sequence $a_1 = 1, a_n = a_{n-1}+\frac{1}{a_{n-1}}$ converge?
Since the function $x+1/x$ is strictly monotonic increasing for all $x>1$, I don't think that the limit converges, but I'm not sure. Can anybody tell me whether the sequence is converging or not?
 A: Assume it converges, then it does so to a limit $L \ge 1$.
Then we have $L = L + \frac{1}{L}$ which is not possible.
A: It is obvious that $a_n\le n$. Thus
$$a_{n+1}\ge a_n+\frac1n.$$
Since $H_n$ diverges, we can conclude that $a_n$ also diverges.
A: Note that $a_n^2 = \left(a_{n-1}+\dfrac{1}{a_{n-1}}\right)^2 = a_{n-1}^2 + 2 + \dfrac{1}{a_{n-1}^2} \ge a_{n-1}^2 + 2$. 
Therefore, $a_n^2 \ge 2n-1$, and thus, $a_n \ge \sqrt{2n-1}$ for all $n \ge 1$. That's enough to show divergence.
A: No. If it were convergent to some $\alpha$, this value would verify 
$$\alpha=\alpha+\frac{1}{\alpha}.$$
A: Since the equation $x= x+\frac1x$ does not have a solution. Therefore $a_n$ does not converges.  Also, since $a_1>0,$ by induction we easily have $a_n>0$ and 
then, $$a_{n+1} -a_n = \frac{1}{a_n}>0$$
which means $(a_n)$ is a strictly increasing and non convergent sequence. So $a_n\to\infty$.  That $a_n$ is unbounded.
A: Since $a_{n+1}=a_n$ + $\frac{1}{a_n}$ and  $a_n$
$\frac{1}{a_n}$ > $0$ we get that $a_{n+1}>a_n$
thus diverging
