# Finding $\lim\limits_{n→∞}n\cos x\cos(\cos x)\cdots\underbrace{\cos(\cos(\cdots(\cos x)))}_{n\text{ times of }\cos}$

Find$$\lim_{n→∞}n\cos x\cos(\cos x)\cdots\underbrace{\cos(\cos(\cdots(\cos x)))}_{n \text{ times of } \cos}.$$

I approximated cos(cosx) to cos x, but i don't think it is the proper approach.
I got answer as 0 on the approximation.
It is clear that it is a 0/0 form, but how can the l's Hopital rule be applied? I tried using the sandwich theorem but I am unable to reach the answer.
I plotted the graph on desmos. But I got the resultant graph covering the entire area. please help me reach the proper answer. Thanks in advanced to all.

• I edited the expression with MathJax. Please check if it's what you wrote.
May 12 '20 at 4:35
• Yes, thanks for that! May 12 '20 at 4:36
• If $x=0$ then $\cos x=1$. $\cos 1\neq 0$ and I don't think that repeating $\cos$ will approach $0$, so I think the limit goes to $\infty$... May 12 '20 at 4:48
• But no constraint is given on x, and we cannot assume x=0 May 12 '20 at 4:51
• It is amazing that nobody mentioned Dottie number May 12 '20 at 5:52

Consider the sequence $$x_n$$ defined by $$x_0 = x$$ and $$x_{n+1} = \cos(x_n)$$. Then the sequence in question is $$a_n = n\prod_{n=1}^\infty x_n.$$ I claim that $$a_n \to 0$$. Here's a sketch of the proof:

1. There is a unique point $$x^* \in [0, 1)$$ such that $$\cos(x^*) = x^*$$ (the fixed point of $$\cos$$).
2. The sequence of fixed point iterates $$x_n$$ converge to $$x^*$$, regardless of the value of $$x$$.
3. The sum $$\sum a_n$$ converges, using the ratio test.
4. The sequence $$a_n$$ converges to $$0$$, using the divergence test.

The hard bit is 2, which I'll leave to last. To prove 1, note that the function $$f(x) = x - \cos(x)$$ is continuous, negative at $$0$$, and positive at $$\pi/2$$, so by the intermediate value theorem, there must be at least one point where $$f(x) = 0$$ in $$[0, \pi/2]$$.

Further, $$f'(x) = 1 + \sin(x) \ge 0$$, meaning that the function is non-decreasing. If $$f$$ had more than one root, then it'd be an interval of roots, which would correspond to an interval of roots in $$f'$$. This is clearly not the case, so there is a unique $$x^*$$ such that $$f(x^*) = 0$$, i.e. $$\cos(x^*) = x^*$$.

The point $$x^*$$ lies in the range of $$\cos$$, i.e. $$[-1, 1]$$, as well as $$[0, \pi/2]$$, so $$x^* \in [0, 1]$$. If $$x^* = 1$$, then $$\cos(x^*) = 1$$, hence $$x^*$$ would have to be an integer multiple of $$2\pi$$, which it is clearly not. Thus, $$x^* \in [0, 1)$$, as claimed.

To prove 3, assuming 2 is proven, consider $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{n}|x_{n+1}| \to x^* < 1,$$ thus the series converges absolutely. Then, 4 follows immediately from this: the terms of a convergent series must tend to $$0$$.

Now, we tackle 2. First, recall the trigonometric identity: $$\cos(x) - \cos(y) = -2\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right).$$ Now, suppose that $$x, y \in [0, 1]$$. Note that $$\frac{x + y}{2} \in [0, 1]$$ and $$\sin$$ is increasing and positive on $$[0, 1] \subseteq [0, \pi/2]$$, hence $$\left|\sin\left(\frac{x + y}{2}\right)\right| = \sin\left(\frac{x + y}{2}\right) \le \sin(1).$$ Also, recall that $$|\sin \theta| \le |\theta|$$ for all $$\theta$$. Hence, assuming still $$x, y \in [0, 1]$$, $$|\cos(x) - \cos(y)| = 2\left|\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)\right| < 2 \cdot \sin(1) \cdot \left| \frac{x - y}{2}\right| = \sin(1)|x - y|.$$

Now, note that $$x_n \in [-1, 1]$$ for $$n \ge 1$$, and since $$\cos$$ is positive over $$[-1, 1]$$, we have $$x_n \in [0, 1]$$ for $$n \ge 2$$. So, for $$n \ge 2$$, we get $$|x_{n+2} - x_{n+1}| = |\cos(x_{n+1}) - \cos(x_n)| \le \sin(1)|x_{n+1} - x_n|.$$ This implies the series $$\sum_{n=2}^\infty (x_{n+1} - x_n)$$ is absolutely summable, as it passes the ratio test (limsup version): $$\left|\frac{x_{n+2} - x_{n+1}}{x_{n+1} - x_n}\right| = \frac{|x_{n+2} - x_{n+1}|}{|x_{n+1} - x_n|} \le \sin(1) \frac{|x_{n+1} - x_n|}{|x_{n+1} - x_n|} = \sin(1) < 1.$$ Therefore, $$\sum_{n=2}^\infty (x_{n+1} - x_n)$$ converges. This is a telescoping series, whose partial sums take the form $$x_n - x_2$$. These partial sums converge, and hence so must $$x_n$$.

Now, because $$x_n$$ converges to some $$L$$, it follows from $$\cos$$ being continuous that $$x_{n+1} = \cos(x_n) \implies L = \cos(L) \implies L = x^*,$$ completing step 2, and the full proof, as necessary.

Let $$x_0$$ denote the root of $$\cos(x) = x$$, then we have $$x_0 \approx 0.739085$$. Define $$\cos_n(x) = \underbrace{\cos(\cos(\cdots(\cos x)))}_{n \text{ times of} \cos}.$$

First, we can prove that for all $$x \in \mathbb{R}$$, \begin{align} \lim_{n \to \infty} \cos_n(x) = x_0.\tag{1} \end{align} From (1), for any given $$x$$, there exists $$N_0(x) > 0$$, such that for all $$k > N_0(x)$$ we have $$|\cos_k(x)| < \frac{x_0 + 1}{2}$$. Therefore, for all $$n > N_0(x)$$, \begin{align} 0 \leq \left|n \prod_{k = 1}^{n} \cos_k(x) \right| &< n \left|\prod_{k = 1}^{N_0(x)} \cos_k(x)\right| \left(\frac{x_0 + 1}{2}\right)^{n - N_0(x)}\\ &= \left|\prod_{k = 1}^{N_0(x)} \cos_k(x)\right| \left(\frac{x_0 + 1}{2}\right)^{-N_0(x)} \cdot n \,\left(\frac{x_0 + 1}{2}\right)^n. \end{align} Since $$\left|\frac{x_0+1}{2}\right| < 1$$, from the squeeze theorem we have $$\lim_{n \to \infty} n \prod_{k = 1}^{n} \cos_k(x) = 0.$$

To prove (1) note that the range of cosine function is $$[-1, 1]$$, and from Lagrange's mean value theorem we have \begin{align} \left|\cos_n(x) - x_0\right| &= \left|\cos(\cos_{n-1}(x)) - \cos(x_0)\right| \\ &\leq \sin(1) \cdot |\cos_{n-1}(x) - x_0| \leq \dots \leq \sin^{n-1}(1) \cdot \left|\cos(x) - x_0\right|. \end{align}

We can say few more things. With the same notations as before: consider the sequence $$x_{n+1} = \cos(x_n)$$ with $$0 < x_0 \leq 1$$, and $$x_*= \cos(x_*)$$.

Heuristically, $$x_n \to x^*$$ very quickly. Thus, $$x_0 x_1 ... x_n$$ behaves like a geometric sequence : $$x_0 x_1 ... x_n \propto x_*^n \simeq 0.73^n \to 0$$

(consequently $$n^k x_0 x_1 ... x_n \to 0$$ for all $$k > 0$$)

To be more precise, consider: $$y_n := \frac1{x_*^{n+1}}x_0 x_1 ... x_n = \frac{x_0}{x_*} \frac{x_1}{x_*} ... \frac{x_n}{x_*}> 0$$

It's better to consider $$\log y_n$$ : $$\log y_n = \sum_{i = 0}^n \log(\frac{x_i}{x_*})$$ This série is really cool. First, the sign of $$(-1)^i \log(\frac{x_i}{x_*})$$ is constant. Indeed, if $$x_n < x_*$$ then $$x_{n+1} > x_*$$ : $$x_{n+1} - x_* = \cos(x_n) - \cos(x_*)$$

In fact, this série is geometric : $$|\log(\frac{x_i}{x_*})| = |\log(x_i) - \log(x_*)| = |\log(\cos(x_{i-1})) - \log(\cos(x_*))| \leq k |x_{i-1} - x_0|$$ with $$k = \displaystyle\sup_{[0, 1]} |\tan| < \infty$$. And, with a Taylor expansion : $$|\log(\frac{x_i}{x_*})| \leq k (\cos(1))^{i-1} |x_1 - x_0|$$

Thus, $$y_n$$ converges.