An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$

Question

I have a sequence $\{r_n\}$ of positive numbers defined recursively by $$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$ for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by induction one can prove

$$\tag{2} r_n^2 \ge 2(n-1)K +r_1^2$$ for all $n$.

Question: Does there exists positive $K_2$ depending on $r_1, K$ such that $$\tag{3} r_n^2 \le 2nK + K_2$$ for all large $n$?

Squaring (1) on both sides give $$ r_n^2 = r_{n-1}^2 + 2K + \frac{K^2}{r_{n-1}^2},$$

If I try to prove (3) using induction, (2) gives

$$ \frac{K^2}{r_{n-1}^2} \le \frac{K^2}{ 2(n-2)K +r_1^2},$$ which is not very helpful, since $\sum \frac{1}{n-2}$ is not summable.

Answer 1

First, reproving the lower bound: we have, for all $n\geq 1$, $$ r_{n-1}^2 +K = r_n r_{n-1} \leq \frac{r_n^2+ r_{n-1}^2}{2} \tag{1} $$ using the AM-GM inequality; from which $$ r_n^2 \geq r_{n-1}^2 + 2K \tag{2} $$ leading to $r_{n}^2 \geq 2K n + r_0^2$ for all $n\geq 0$.

Then, the upper bound: We know, using the lower bound, that there exists $N\geq 1$ such that, for all $n\geq N$, $$ r_n^2 \geq 2Kn $$ so we get now use this to get $$ r_n = r_{n-1}+\frac{K}{r_{n-1}} \leq r_{n-1}+\frac{\sqrt{K}}{\sqrt{2(n-1)}} \tag{3} $$ so that $$ r_n - r_1 = \sum_{k=2}^n (r_k - r_{k-1}) \leq \sum_{k=2}^n \sqrt{\frac{K}{2(k-1)}} = \sqrt{\frac{K}{2}}\sum_{k=1}^{n-1} \frac{1}{\sqrt{k}} \leq \sqrt{\frac{K}{2}}\int_{0}^{n-1} \frac{dx}{\sqrt{x}} = \sqrt{2K(n-1)} $$ and so $(r_n-r_1)^2 \leq 2Kn$. Now, since $r_n \to \infty$, for every $\varepsilon>0$ there exists some $n_\varepsilon\geq 1$ such that $(r_n-r_1)^2 \geq (1+\varepsilon/2)^{-1}r_n^2$ for all $n\geq n_\varepsilon$.

Conclusion. For all $\varepsilon>0$, there exists $n_\varepsilon$ such that, for $n\geq n_\varepsilon$, $$ \boxed{ 2Kn + r_0^2 \leq r_n^2 \leq (2+\varepsilon)Kn + r_1^2 } $$ (the lower bound actually holds for all $n\geq 0$).

Answer 2

tl;dr: You cannot prove the upper bound you want, as the true growth of the sequence is "slightly more" than $2Kn$: $$ r_n^2 = 2Kn + \frac{K}{2}\log n + O(1) $$

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First, for the sake of "elegance", set $a_n := \frac{r_n}{\sqrt{K}}$ for all $n\geq 0$, so that the recurrence relation becomes $$ a_n = a_{n-1} + \frac{1}{a_{n-1}}, \qquad n\geq 1 \tag{1} $$

Squaring both sides, we get $a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2}$, or, equivalently, $$ a_n^2 - a_{n-1}^2 = 2 + \frac{1}{a_{n-1}^2} \tag{2} $$

Summing (2) from $1$ to $n$, we get $$ a_n^2 - a_0^2 = 2n + \sum_{k=0}^{n-1}\frac{1}{a_k^2} \tag{3} $$ and in particular $a_n^2 \geq 2n + a_0^2$ for all $n\geq 0$. We also have, analogously to (3), that $$ a_n^2 - a_1^2 = 2(n-1) + \sum_{k=1}^{n-1}\frac{1}{a_k^2} $$ and using our lower bound $a_k^2 \geq 2k$, we get $$ a_n^2 \leq 2n-2+a_1^2 + \frac{1}{2}\sum_{k=1}^{n-1}\frac{1}{k} \geq 2n + \frac{1}{2} \log n +a_1^2 $$

At this point, we have shown, for all $n\geq 1$, $$ \boxed{2n + a_0^2 \leq a_n^2 \leq 2n + \frac{1}{2} \log n +a_1^2} \tag{4} $$ This is great, but let us not stop there. Plugging (4) back into (3), we have $$ a_n^2 - a_0^2 \geq 2n +\frac{1}{a_0^2} + \sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n +a_1^2} \geq 2n + \frac{1}{2}\log n + C \tag{5} $$ for some constant $C\in\mathbb{R}$ which depends on $a_0,a_1$ only (and therefore on $a_0$ only, since $a_0$ determines $a_1$): where we used that $$ \sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n + a_1^2} = \frac{1}{2}\sum_{k=1}^{n-1} \left(\frac{1}{n}\cdot \frac{1}{1+\frac{\log n + a_1^2}{n}}\right) = \frac{1}{2}\sum_{k=1}^{n-1} \frac{1}{n} - \sum_{k=1}^{n-1}\left(\frac{\log n}{2n^2} + o\!\left(\frac{\log n}{n^2}\right)\right) $$ and the second series is convergent.

Overall, we have therefore established that there exist a constant $C\in\mathbb{R}$ (depending on the initial term $a_0$ only) such that, for all $n\geq 1$, $$ \boxed{2n + \frac{1}{2}\log n + C \leq a_n^2 \leq 2n + \frac{1}{2} \log n + a_1^2} \tag{6} $$ that is, $a_n^2 = 2n + \frac{1}{2}\log n + O(1)$.