# 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?

• Before trying to find a formal proof, it helps to get some intuition. For that, I suggest sketching by hand the web diagram of your function. Aug 4 '14 at 18:44
• Once you showed that it is not converging, you can find the speed by considering the differential equation $f'=1/f$ Aug 4 '14 at 18:46
• No, but at large values it does become linear! Oct 19 '20 at 14:25

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.

• How did you get that $a_n^2 \geq 2n-1$? Sep 11 '18 at 18:47

No. If it were convergent to some $\alpha$, this value would verify $$\alpha=\alpha+\frac{1}{\alpha}.$$

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.

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.

• What does $H_n$ stand for?
Mar 23 '17 at 1:51
• @Vlad It is the harmonic series, defined as $H_n=\sum_{k=1}^n 1/k$. Mar 28 '17 at 11:23

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.

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