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Put $ a_1 = 1$ and $a_{n+1} = a_n +\frac{1}{a_n}$.

1) Main question: what can we say about the speed of convergence of $\frac{a_n}{\sqrt{n}}$ to its limit?

2) Additional question: is there a simple way to find the limit of $\frac{a_n}{\sqrt{n}}$?

More formally: $\lim_{n \to \infty }\frac{a_n}{\sqrt{n}} = \sqrt{2}$ (if there's no mistake below), what can we say about $\frac{a_n}{\sqrt{n}} - \sqrt{2}$, e.g. does $\lim_{n \to \infty} (a_n - \sqrt{2n})$ exist and can we find it in an explicit form? From direct computation in Python it follows (see above) that $$\text{Hypothesis:} \quad \lim_{n \to \infty} (a_n - \sqrt{2n}) = 0$$ is true.

What is obvious? It's easy to see that $a_n$ is strictly increasing and tends to $\infty$ (Indeed, otherwise $a_n$ is bounded and hence $a_n \to a < \infty$. Thus $a = \lim a_{n+1} = \lim (a_n +\frac{1}{a_n}) = a +\frac{1}{a} \Longrightarrow a = a+\frac{1}{a}$).

What else do we know? If I'm not mistaken $\big[\sqrt{2n+\frac94} - \frac12\big] \le a_n \le \sqrt{2n}(1+o(1))$ for all $n > 2$ (the proof is above) and hence $\lim_{n \to \infty} \frac{a_n}{\sqrt{n}} = \sqrt{2}$. Some numerical experiments are presented in the end of the question.

What are the difficulties? The upper bound for $a_n$ was obtained using the asymptotics of the sums $$\sum_{n=1}^{\infty} \frac{1}{[f(n)]}$$ where $f(n) = \sqrt{2n+\frac94} - \frac12$ and $[x]$ is an integer part of $[x]$. There are good estimates (using intergal approximation) of $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ for good functions $f$, but it looks like there are no good estimates of $\sum_{n=1}^{\infty} \frac{1}{[f(n)]}$.

Proofs:

Lemma1. $a_n \ge \sqrt{2n+\frac94} - \frac12$ if $n = \frac{k(k+1)}2-1 \ge 2$.

Proof. Lets prove that $a_{\frac{k(k+1)}2 - 1} \ge k$ for $k \ge 2$. Put $m_k = \frac{k(k+1)}2-1$. We will use induction and suppose the opposite. We have: $a_{m_k} \ge k$ and $a_{m_{k+1}} < k+1$. Hence $$ k \le a_{m_k} < a_{m_k + 1 } < a_{m_k + 2 } < \ldots < a_{m_{k+1}} < k+1,$$ $$ a_{m_k+i+1} - a_{m_k+i} = \frac{1}{a_{m_k+i}} > \frac{1}{k+1}$$ Summing l.h.s. or the last inequality we get $$ a_{m_{k+1}} - a_{m_{k}} > \frac{1}{k+1} + \frac{1}{k+1} + \ldots + \frac{1}{k+1} = \frac{m_{k+1}-m_k}{k+1}.$$ But $a_{m_{k+1}} - a_{m_{k}} < (k+1)- k = 1$. Thus $$ 1 > \frac{m_{k+1}-m_k}{k+1} \Longrightarrow m_{k+1}-m_k < k+1 \Longrightarrow \frac{(k+1)(k+2)}2 - \frac{k(k+1)}2 < k+1 \Longrightarrow $$ $$ \Longrightarrow \frac{(k+2)}2 - \frac{k}2 < 1 \Longrightarrow 1 < 1.$$ We got a contradiction. Thus $a_{m_{k+1}} \ge k+1$. Lemma is proved.

Lemma2: $a_n \ge \big[\sqrt{2n+\frac94} - \frac12\big]$ for all $n \ge 2$.

Proof. Put $u_n = \big[\sqrt{2n+\frac94} - \frac12\big]$. We have $$ u_n \le \sqrt{2n+\frac94} - \frac12 \Longrightarrow (u_n + \frac12)^2 \le 2n+\frac94 \Longrightarrow n \ge \frac{u_n(u_n+1)}{2}-1 $$ Thus $$ a_n \ge a_{\frac{u_n(u_n+1)}{2}-1} \ge u_n$$ by lemma 1.

Lemma3: $\frac{a_n}{\sqrt{n}} \to \sqrt{2}$ as $n\to \infty$.

Proof. According to lemma 2 it's sufficient to show that $\overline{\lim}_{n \to \infty} \frac{a_n}{\sqrt{n}} \le \sqrt{2}$.

Put $u_n = \big[\sqrt{2n+\frac94} - \frac12\big]$. We have $$ a_{j+1} - a_j = \frac1{a_j} \le \frac{1}{u_j}$$ It follows that $$ a_n = \sum_{j=3}^{n-1} (a_{j+1}-a_j) + a_2 \le 2 + \sum_{j=3}^{n-1} \frac{1}{u_j}$$ As $u_j \sim \frac{1}{\sqrt{2j}}$ hence $$\sum_{j=3}^{n-1} \frac{1}{u_j} \sim \sum_{j=1}^{n} \frac{1}{\sqrt{2j}} \sim \int_{1}^n \frac{dx}{\sqrt{2x}} \sim \sqrt{2n}.$$ Thus $\overline{\lim}_{n \to \infty} \frac{a_n}{\sqrt{n}} \le \sqrt{2}$.

Numerical experiments.

Let's compute $b_n = a_n - \sqrt{2n}$ for first values of $n$ in Python. We get $b_{1000}=0.035513542064180115, b_{2000}=0.0278533072346292, b_{3000}=0.02405133309876817, b_{4000}=0.021633453406735725, b_{5000}=0.019907583619357183, b_{6000}=0.018589261056305872, b_{7000}=0.017536096833907777, b_{8000}=0.01666749243514687, b_{9000}=0.015933785448964954, b_{10000}=0.015302406276020974$

Code:

tmp = 1

for i in range(1,11000):

if i % 1000 == 0:

    print('i = ', i, 'a_i/sqrt_i - (2*i)^(0.5) ',  tmp - (2*i)**0.5)

tmp = tmp + 1/tmp

It looks like $b_{n} > b_{n+1} > 0$ and $b_n \to 0$ as $n \to \infty$.

Addition (an idea that might be helpful): there's the next idea which looks like it may be useful but I don't know how to apply it. An equation $a_{n+1}-a_{n} = \frac{1}{a_n}$ looks like an ODE $y' = \frac{1}{y(t)}$, Solving ODE we get $2y'y = 2 + c \Longrightarrow y^2 = 2t + c\Longrightarrow y = \sqrt{2t + c}$. Then maybe we may get properties of $a_n$ from the solution $y(t) = \sqrt{2t + c}$ if we estimate the precision of approximation of ODE by our linear recurrence relation. I am not an expert in the field of such approximations. If you think that idea with $y(t)$ is useful please give a hint how to bring this idea to mind.

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4 Answers 4

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Clearly $\{a_n\}$ is increasing. Suppose $\{a_n\}$ is bounded. Then $\lim_{n\to\infty} a_n=a$ exists. So one has $$ a=a+\frac1a$$ which is impossible. So $\{a_n\}$ is unbounded and hence $a_n\to\infty$ as $n\to\infty$. By Stolz' Theorem, one has \begin{eqnarray} \lim_{n\to\infty} \frac{a_n^2}{n}&=&\lim_{n\to\infty} \frac{a_{n+1}^2-a_n^2}{(n+1)-n}\\ &=&\lim_{n\to\infty} (a_{n+1}-a_n)(a_{n+1}+a_n)\\ &=&\lim_{n\to\infty} \frac{2a_n+\frac1{a_n}}{a_n}\\ &=&2+\lim_{n\to\infty} \frac{1}{a_n^2}=2. \end{eqnarray}

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  • $\begingroup$ $\displaystyle +1$. Nice job. $\endgroup$ Jan 24, 2023 at 17:00
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    $\begingroup$ @FelixMarin Thank you! $\endgroup$
    – xpaul
    Jan 24, 2023 at 20:06
  • $\begingroup$ Thanks! I have never seen Stolz' Theorem before. Interesting. Never too old to learn :-) $\endgroup$ Jan 24, 2023 at 21:20
  • $\begingroup$ @HandeBruijn, you are welcome. $\endgroup$
    – xpaul
    Jan 24, 2023 at 21:45
  • $\begingroup$ @xpaul Thank you! $\endgroup$ Jan 25, 2023 at 8:16
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The analysis concerning the ratio $a_n/\sqrt{2n}$ can be shortened. We have $$a_{n+1}^2-a_n^2=(a_{n+1}-a_n)(a_{n+1}+a_n)={a_{n+1}\over a_n}+1\ge 2$$ Hence $a_n^2\ge 2n-1.$ Furthermore $$a_{n+1}-a_n={1\over a_n}\le {1\over \sqrt{2n-1}}$$ Thus $$a_{n+1}\le 1+\sum_{k=1}^{n}{1\over \sqrt{2k-1 }}=2+\sum_{k=2}^n{1\over \sqrt{2k-1}}\\ \le 2+\sum_{k=2}^{n}[\sqrt{2k-1}-\sqrt{2k-3}]=1+\sqrt{2n-1}$$

Remark By considering $a_n^2$ gives $$a_{n+1}^2=a_n^2+2+{1\over a_n^2}$$ Similar analysis gives better estimates $2n+2\le a_n^2\le 2n+2+{\ln n\over 2}$ mentioned in one of the answers, although the lower estimate is wrong for small indices, e.g. $n=1,2.$ It should be replaced by $a_n^2\ge 2n,$ $n\ge 2.$

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  • $\begingroup$ @ RyszardSzwarc, Thank you! $\endgroup$ Jan 25, 2023 at 8:17
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We know (see Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$) that $$2n+2\le a_n^2\le 2n+2+\frac{\ln n}{2}.$$ Hence, $$\sqrt{2n+2}-\sqrt{2n}\le b_n\le \sqrt{2n+2+\frac{\ln n}{2}}-\sqrt{2n}.$$ Using equality $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ and sandwich theorem we get that $b_n\to 0$. Moreover, $b_n>0$.

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  • $\begingroup$ Thank you! I didn't expect that it was already almost solved. $\endgroup$ Jan 23, 2023 at 18:19
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It's rather easy to prove that $\bbox[navy,8px]{\color{white}{a_n=\sqrt{2n} + \frac{\ln n}{(4\sqrt2)\sqrt n} + o\big(\frac{\ln n}{\sqrt n}\big)}}$, as $n\to\infty$ (assuming the results already proven in the question). Let's put $u_n=a_n^2-2n$ for $n\geqslant0$.

Then $u_{n+1}=a_{n+1}^2-2(n+1) =a_n^2+2+a_n^{-2}-2n-2 =u_n + a_n^{-2}$.

Then $u_n=u_1+\sum^{n-1}_{k=1}(u_{k+1}-u_k) =u_1+\sum^{n-1}_{k=1} a_k^{-2}$.

Now, since $a_k^{-2}\sim \frac1{2k}$ as $k\to\infty$, we have $\sum^{n-1}_{k=1} a_k^{-2} \sim \sum^{n-1}_{k=1} \frac{1}{2k} \sim \frac12\ln n$ (as $n\to\infty$).

We thus have $u_n\sim\frac12\ln n$, which lastly provides: \begin{eqnarray*} a_n&=&\sqrt{u_n+2n}=\sqrt{2n}\sqrt{1+\frac{u_n}{2n}}\\ &=&\sqrt{2n}\sqrt{1+\frac{\ln n}{4n}+o\Big(\frac{\ln n}{n}\Big)}\\ &=&\sqrt{2n}\bigg(1 + \frac{\ln n}{8n} + o\Big(\frac{\ln n}{n}\Big)\bigg)\\ &=&\sqrt{2n}+\frac{\ln n}{(4\sqrt2)\sqrt n} + o\bigg(\frac{\ln n}{\sqrt n}\bigg)\cdot \end{eqnarray*}

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  • $\begingroup$ It's interesting, thank you! $\endgroup$ Jan 26, 2023 at 16:54
  • $\begingroup$ You're welcome. $\endgroup$ Jan 26, 2023 at 16:58

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