Limit of the sequence $u_{n+1}=2^{n+1}\arctan \left(\frac{u_{n}}{2^{n+1}}\right)$

Question

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 ?.

Answer 1

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 $u_0$.

Answer 2

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.

Answer 3

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.

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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.

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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 $f(t) \geqslant 0 \quad \forall t > 0$.

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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.

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