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I am interested in calculating the limit of the sequence $$ u_{n + 1} = 2^{n + 1} \arctan\left(\dfrac{u_{n}}{2^{n + 1}}\right) \,,\qquad\left(\ u_{0} > 0\,\right) $$ By the inequality $\arctan\left(x\right) < x,\ \mbox{if}\quad x > 0\ ,$ we can see that:

  • $u_{n}$ is strictly decreasing and $0 < u_{n} < u_{0}$, thus the sequence converges.
  • As $\arctan\left(x\right) \approx x$ ( if $x$ approaches to $0$ ) and thus $u_{n+1}\approx u_{n}$ ( if $n$ is big ), it converges very slowly.
  • ( a $\tt Python$ script shows that we get a decrease of $0.01$ in $1000$ steps ! ).
    By the fixed point theorem, we can show that
    $$ \dfrac{u_{n}}{2^{n}} \to 0 $$ But it is too little to infer that the limit of $u_{n}$ is zero$\ldots$

Can anyone have an idea, how to prove it ?.

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  • $\begingroup$ a decreasing bounded from below sequence has a clear limiting behavior :-) $\endgroup$
    – Math-fun
    Commented Jun 3 at 13:15
  • $\begingroup$ Yes, but how do you know that the limit is not a little beat above 0, a value reached asymptotically? $\endgroup$
    – user472228
    Commented Jun 3 at 13:48
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    $\begingroup$ by definition of the limit, the sequence can be as a close as one wishes to zero: this makes zero the limit :-) $\endgroup$
    – Math-fun
    Commented Jun 3 at 14:12
  • $\begingroup$ @Math-fun ok, but here it is not true that $u_n$ can be as close to $0$ as we wish. For me it even seems it won't be. I have run some code that showed that it seems to have limit different from $0$ $\endgroup$ Commented Jun 3 at 14:24
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    $\begingroup$ If you start with the sequence $a_{n+1} = (1/2)\arctan (a_n)$ and put $b_n = 2^{n+1} a_n $ then $b_{n+1} = 2^{n+1} \arctan(b_n/2^{n+1})$ is the same as your sequence. Therefore I think this is a dupe of several old questions e.g. math.stackexchange.com/questions/3246828/…, math.stackexchange.com/questions/2823644/…, math.stackexchange.com/questions/4518011/… $\endgroup$ Commented Jun 3 at 19:06

3 Answers 3

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As you noticed, $u_n \rightarrow l(u_0)$ for some limit $l(u_0)$ between $0$ and $u_0$, depending only on $u_0$. You can get a better estimation of $\arctan$ near $0$ than simply $\arctan(x) \sim x$. Indeed, if you set $f(x) = \arctan(x) - x + \frac{x^3}{3}$ (it comes from the Taylor development of $\arctan$ at $x = 0$ at order $4$), you get that, $$ f'(x) = \frac{1}{x^2 + 1} - 1 + x^2 = \frac{1}{x^2 + 1}(-1 + (x^2 + 1)(x^2 - 1)) = \frac{x^4}{x^2 + 1} \geqslant 0, $$ thus $f$ increases so for all $x \geqslant 0$, $f(x) \geqslant f(0) = 0$ $\textit{i.e.}$ $\arctan(x) \geqslant x - \frac{x^3}{3}$. Using this and $\arctan(x) \leqslant x$ with $x = \frac{u_n}{2^{n + 1}}$, we deduce that for all $n$, $$ \frac{u_n}{2^{n + 1}} - \frac{u_n^3}{3 \cdot (2^{n + 1})^3} \leqslant \arctan\left(\frac{u_n}{2^{n + 1}}\right) \leqslant \frac{u_n}{2^{n + 1}}, $$ and after multiplying by $2^{n + 1}$, $$ u_n - \frac{u_n^3}{12 \cdot 4^n} \leqslant u_{n + 1} \leqslant u_n. $$ In particular, since $u_n$ is bounded by $u_0$, $$ u_n - u_{n + 1} \leqslant \frac{u_n^3}{12 \cdot 4^n} \leqslant \frac{u_0^3}{12 \cdot 4^n}, $$ therefore, $$ u_n = u_0 + \sum_{k = 0}^{n - 1} u_{k + 1} - u_k \geqslant u_0 - \sum_{k = 0}^{n - 1} \frac{u_0^3}{12 \cdot 4^k} = u_0 - \frac{u_0^3}{12}\frac{1 - 4^{-n}}{1 - 4^{-1}} = u_0 - \frac{u_0^3}{3}(1 - 4^{-n}). $$ When $n \rightarrow +\infty$, we deduce that $u_0 - \frac{u_0^3}{3} \leqslant l(u_0) \leqslant u_0$. It implies that $l(u_0) \sim u_0$ when $u_0 \rightarrow 0$. In particular, it implies that $l(u_0) > 0$ when $u_0 > 0$ is small. Since $l$ is clearly an increasing function of $u_0$, we deduce that for all $u_0 > 0$, the limit $l(u_0)$ is positive too.

Using the same kind of procedure with the sharper bounds, $$ \sum_{k = 0}^{2N + 1} (-1)^k\frac{x^{2k + 1}}{2k + 1} \leqslant \arctan(x) \leqslant \sum_{k = 0}^{2N} (-1)^k\frac{x^{2k + 1}}{2k + 1}, $$ for all fixed integer $N$ and all $x \geqslant 0$, you can probably get a better approximation of the function $l$ near $0$, or even an exact expression when $N \rightarrow +\infty$ as a series in $u_0$.

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Let's rewrite as $$u_n=2^n\arctan\left(\frac12\arctan\left(\frac12\arctan\left(\ldots\arctan\left(\frac12u_0\right)\ldots\right)\right)\right),$$ where the compositions are $n$ times. Note let $u_n=2^nx_n$. Clearly, with $x_n=\arctan\left(\frac12x_{n-1}\right)$, we obtain $\lim_{n\to\infty}x_n=0$ (decreasing bounded, limit exists, etc). Now since $$\lim\frac{1}{x_n^2}-\frac4{x_{n-1}^2}=\frac23, \qquad (*)$$ a sequence of the form $y_n=\frac1{x_{n}^2}$ diverges at the rate of $$\frac23\left(1+4+4^2+\cdots4^n\right)\sim\frac23\frac{2^{2n+2}{}}{3}.$$ Therefore, $$x_n\sim\frac{3}{\sqrt2}2^{-n-1},$$ and hence the desired limit is $$\frac3{\sqrt2}.$$

Note on $(*)$: this is to pin down the higher order convergence of the the nested sequence to zero and is a little helpful trick by looking at the easier to follow divergence of the inverse of the sequence. Note that the dominant term of $\frac1{\left(\arctan\frac{x}{2}\right)^2}$ when $x$ is close to zero is $\frac4{x^2}$, and the next term is $\frac23$, therefore, we can get a hand on how does $x_n$ vanishes by looking at the higher order term difference which I formed.

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    $\begingroup$ I don't think it is a limit, since we know $u_n$ is decreasing and by taking small $u_0$ it can reach pretty small numbers (checked computationally). $\endgroup$ Commented Jun 4 at 9:51
  • $\begingroup$ @RupertRybka thank you for the comment: do you see a mistake in my claculations? $\endgroup$
    – Math-fun
    Commented Jun 4 at 15:15
  • $\begingroup$ I am not familiar with that kind of approach so, I think I can't help much there $\endgroup$ Commented Jun 4 at 15:19
  • $\begingroup$ $1/x_n^2 - 4/x_{n - 1}^2 \rightarrow 2/3$ seems to come form nowhere, how did you get it ? $\endgroup$
    – Cactus
    Commented Jun 5 at 9:08
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    $\begingroup$ But still if you took $u_0 = 1$ you have $u_n$ decreases so it cannot have limit $\frac{3}{\sqrt{2}}$ $\endgroup$ Commented Jun 5 at 15:15
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In my attempt I will prove the following inequality: $$ \boxed{u_n \geqslant \arctan(u_0)} $$ which implies, that the limit of that sequence is not $0$ and gives us some nice lower bound for that sequence.


In my approach I will use some results derived by @Cactus and @Math-fun in their solutions.

Let's at first notice that: $$ u_n = 2^n \arctan\left(\frac{1}{2}\arctan\left(\dots\left(\frac{1}{2}\arctan\left(\frac{u_0}{2}\right)\right)\right)\right) $$ where we compose $\arctan$ function $n$ times. This motivates us to denote sequence $\{x_n\}$, where $x_0 = u_0$ and $x_n = \arctan(\frac{x_{n-1}}{2})$. This gives us relation $u_n = 2^nx_n$. Values of $x_n$ are for sure positive.

We have the identity $\arctan(x) \leqslant x$ that holds for all positive real values of $x$. This allows us to write: $$u_0 \geqslant \arctan(u_0)$$

Now let's notice that: $$ x_n = \arctan\left(\frac{x_{n-1}}{2}\right) \leqslant \frac{x_{n-1}}{2} $$ If we apply that identity $n-1$ times, we will get: $$ x_n \leqslant \frac{x_1}{2^{n-1}} \implies x_n^3 \leqslant \frac{x_1^3}{2^{3n - 3}} \implies 2^{3n-3}x_n^3 \leqslant x_1^3 \qquad (1) $$ which we will use later.


To prove the result I will use induction with the following (stronger) claim: $$ u_n = \boxed{2^nx_n \geqslant \arctan(u_0) + \frac{x_1^3}{9\cdot2^{2n-3}}} > \arctan(u_0) $$ Base case for $n = 1$.

We have: $$ 2x_1 \geqslant \arctan(u_0) + \frac{2\cdot x_1^3}{9} \iff 2\arctan\left(\frac{x_0}{2}\right) \geqslant \arctan(x_0) + \frac{2}{9}\cdot\arctan\left(\frac{x_0}{2}\right)^3 $$ Replcaing $\frac{x_0}{2}$ with $t$ tells us, it is sufficient to show, that: $$ f(t) = 2\arctan(t) - \arctan(2t) - \frac{2}{9}\cdot\arctan(t)^3 \geqslant 0 \qquad \forall t > 0 $$

We have: $$ \lim_{t\to \,+\infty} f(t) = \pi - \frac{\pi}{2} - \frac{2}{9} \cdot \frac{\pi^3}{8} = \frac{18\pi - \pi^3}{36} > 0 \qquad \text{(because $\pi^2 < 18$)} $$ This implies, that: $ \exists t_0\forall t > t_0: f(t) > 0 $. So after some point $f(t)$ is positive so we can pick some $t_1 > t_0$ and then $f(t_1) > 0$. In addition we have $f(0) = 0$.We know that, $f(t)$ is continuous, so it has some minimum in the interval $[0, t_1]$. With what was said, the minimum is either $0$ or some value $f(t)$, where $t \in (0, t_1)$. But $f(t)$ is also differentiable, so if it has a minimum in $(0, t_1)$, then in that point we have $f'(t) = 0$.

Now I will show that: $f'(t) = 0 \implies f(t) \geqslant 0$.

We have: $$ f'(t) = \frac{2}{3\cdot(1 + t^2)(1 + 4t^2)}(9t^2 - \arctan(t)^2(1 + 4t^2)) $$ Now: $$ f'(t) = 0 \iff 9t^2 - \arctan(t)^2(1 + 4t^2) = 0 \iff \arctan(t) = \frac{3t}{\sqrt{4t^2 + 1}} $$ Moreover we can observe that, if this equation has a solution, then $t > 1$. By way of contraditction, take $t \leqslant 1$, then $\arctan(t) \leqslant \frac{\pi}{4} < 1$. In addition we have $\arctan(t) \leqslant t$. So now: $$ 9t^2 - \arctan(t)^2(1 + 4t^2) \geqslant 9t^2 - t^2 - 4t^2 = 4t^2 > 0 $$ and this is the contradition.

Now let's take $T > 1$ such that $\arctan(T) = \frac{3T}{\sqrt{4T^2 + 1}}$ (if such $T$ does not exist, then we immedietely have $f(t) \geqslant 0$ $\forall t > 0$)$^{(*)}$. Additionally I will use the fact $\arctan(t) \leqslant \frac{\pi}{2}$. $$ f(T) \geqslant 2\arctan(T) - \frac{\pi}{2} - \frac{2}{9}\arctan(T)^3 = h(T) $$ I can use equality $\arctan(T) = \frac{3T}{\sqrt{4T^2 + 1}}$ to obtain: $$ h(T) = \frac{6T}{\sqrt{4T^2 + 1}} - \frac{\pi}{2} - \frac{6T^3}{(4T^2 + 1)\sqrt{4T^2 + 1}} $$ $$ h(T) \geqslant 0 \iff 6T - \frac{6T^3}{(4T^2 + 1)} \geqslant \frac{\pi}{2}\sqrt{4T^2 + 1} $$ LHS and RHS of the last inequality are both positive for $T > 0$, so we can square them. Then we can multiply both sides by $(4T^2 + 1)^2$ and after rearranging terms we will obtain: $$ (324 - 16\pi^2)T^6 + (216 - 12\pi^2)T^4 + (36 - 3\pi^2)T^2 - \frac{\pi^2}{4} \geqslant 0 $$ Let's substitute $z = T^2 \quad(T > 1 \implies z > 1)$. $$ g(z) = (324 - 16\pi^2)z^3 + (216 - 12\pi^2)z^2 + (36 - 3\pi^2)z - \frac{\pi^2}{4} \geqslant 0 $$ Now: $$ g'(z) = 3(324 - 16\pi^2)z^2 + 2(216 - 12\pi^2)z + (36 - 3\pi^2) $$ From inequality $\pi^2 < 10$ we can deduce that all coefficients of that quadratic function are positive, and thus it is positive for $z > 0$, so $g(z)$ is increasing.

For $z = 1$ we have: $$ g(1) = 576 - \frac{125\pi^2}{4} > 0 $$ And since $g(z)$ is increasing for $z > 0$, we have $g(z) > g(1) > 0$ for $z > 1$.

Therefore we have:

$$ f(T) \geqslant h(T) > 0 $$

And thus we have $f(t) \geqslant 0 \quad \forall t > 0$.


Inductive step.

I will use inequality $\arctan(x) \geqslant x - \frac{x^3}{3}$ here.

We have: $$ \begin{align} 2^{n+1}x_{n+1} = 2^{n+1}\arctan\left(\frac{x_n}{2}\right) \geqslant 2^{n+1} \left(\frac{x_n}{2} - \frac{x_n^3}{3 \cdot 2^3}\right) = \\ = 2^nx_n - 2^{n-2}\frac{x_n^3}{3} \geqslant \arctan(u_0) + \frac{x_1^3}{9\cdot2^{2n-3}} - 2^{n-2}\frac{x_n^3}{3} \end{align} $$ It suffices to show now, that: $$ \frac{x_1^3}{9\cdot2^{2n-3}} - 2^{n-2}\frac{x_n^3}{3} \geqslant \frac{x_1^3}{9\cdot2^{2n-1}} $$ $$ \frac{x_1^3}{9\cdot2^{2n-3}} - \frac{x_1^3}{9\cdot2^{2n-1}} \geqslant 2^{n-2}\frac{x_n^3}{3} $$ $$ \frac{x_1^3}{3\cdot2^{2n-1}}\geqslant 2^{n-2}\frac{x_n^3}{3} $$ $$ x_1^3\geqslant 2^{3n-3}x_n^3 $$ but this is exactly $(1)$. So the proof is finished.


$^{(*)}$ as noticed by @user472228 such $T$ in fact exists and $ T\approx 13.91$

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  • $\begingroup$ How did you discover your 'stronger claim' ? $\endgroup$
    – user472228
    Commented Jun 5 at 22:08
  • $\begingroup$ I wanted to have something a little bit larger than $\arctan(u_0)$ because otherwise it would be pretty hard to use induction without some extra term. Then it was just some sort of play to get to it motivated by inequality used in inductive step and inequality $(1)$ $\endgroup$ Commented Jun 6 at 6:12
  • $\begingroup$ The most difficult part was in fact base case, because first of all I tried to work with expression with $u_0^3$ rather than with $x_1^3$ but it was failing. I was graphing those functions mentioned in base case in the online programme to check wheteher it makes sense, that mentioned expression is positive $\endgroup$ Commented Jun 6 at 6:16
  • $\begingroup$ In general it can prove helpful to introduce some stronger claim for your induction that depends on the actual variable (like here we had that part $\frac{1}{2^{2n-3}}$), because it might help you to go through inductive step more easily. $\endgroup$ Commented Jun 6 at 6:19
  • $\begingroup$ Thank you very much for your invaluable contribution and your clear comments! $\endgroup$
    – user472228
    Commented Jun 7 at 9:54

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