# Given $a_1 = -1$ and $a_{n+1} = e^{\arctan(a_n)}-1$ find $c$ such that $a_n \sim \frac cn$, $n\to+\infty$

Given $$a_1 = -1$$ and $$a_{n+1} = e^{\arctan(a_n)}-1$$ find $$c$$ such that $$a_n \sim \frac cn$$, $$n\to+\infty$$

I proved that this series converges to $$0$$, but I was unable to do the question in the title. In my book the following solution is given. Using Stolz–Cesàro theorem, we get:

From which we conclude that $$c = -2$$. I see that in this solution we used Taylor expansions of $$e^{\arctan(a_n)}$$ and $$(1 + \frac 12 \arctan(a_n) + O(\arctan^2(a_n))^{-1}$$. I just don't get it where did $$a_n^{-1}$$ go and I am not sure if I am right in my reasoning why $$(1 + \frac 12 \arctan(a_n) + O(\arctan^2(a_n))^{-1} = 1 - \frac 12 \arctan(a_n) + O(\arctan^2(a_n))$$.

For my second question, using Taylor expansion we get that $$(1 + \frac 12 \arctan(a_n) + O(\arctan^2(a_n))^{-1} = 1 - (\frac12\arctan(a_n) + O(\arctan^2(a_n)) + O((\frac12\arctan(a_n) + O(\arctan^2(a_n))^2)$$.

Further $$O((\frac12\arctan(a_n) + O(\arctan^2(a_n))^2) = O(\frac14 \arctan^2(a_n) + \arctan(a_n)O(\arctan^2(a_n) + (O(\arctan^2(a_n))^2)$$

Using the facts that:

$$O(f) + O(f) = O(f)$$

$$\frac 14 \arctan^2(a_n) = O(\arctan^2(a_n))$$

$$\arctan(a_n)O(\arctan^2(a_n)) = O(\arctan^2(a_n))$$

$$(O(\arctan^2(a_n))^2 = O(\arctan^2(a_n))$$

$$O(O(f)) = O(f)$$

we get that $$O(\frac 14 \arctan^2(a_n) + \arctan(a_n)O(\arctan^2(a_n)) + (O(\arctan^2(a_n))^2) = O(\arctan^2(a_n))$$. I am not sure if I made any mistakes here.

• To make it look nicer (as well as for other common functions), you should try using \arctan instead of just writing arctan. Commented Jun 21, 2023 at 15:59

It seems they are using that for $$a_n\to 0$$

$$\arctan a_n=a_n-\frac13 a_n^3+O\left(a^5_n\right)$$

and then

$$a_n^{-1} =\arctan^{-1} a_n+O\left(\arctan a_n\right)\tag 1$$

therefore

$$\left(1+\arctan a_n +\frac12 \arctan^2 a_n +O\left(\arctan^3 a_n\right)-1\right)^{-1}-a_n^{-1}$$

$$\left(\arctan a_n +\frac12 \arctan^2 a_n +O\left(\arctan^3 a_n\right)\right)^{-1}-\arctan^{-1} a_n+O\left(\arctan a_n\right)$$

$$\arctan^{-1} a_n \left(\left(1+\frac12 \arctan a_n+O\left(\arctan^2 a_n\right)\right)^{-1}-1+O\left(\arctan^2 a_n\right)\right)$$

$$\arctan^{-1} a_n\left( \left(1-\frac12 \arctan a_n+O\left(\arctan^2 a_n\right)\right)-1+O\left(\arctan^2 a_n\right)\right)$$

$$\arctan^{-1} a_n\left(-\frac12 \arctan a_n+O\left(\arctan^2 a_n\right)\right)$$

$$-\frac12 +O\left(\arctan a_n\right) \to -\frac12$$

As an alternative we could use that

$$\frac1{e^{\arctan a_n}-1}- \frac1{a_n}=\frac1{e^{\arctan a_n}-1}- \frac1{\arctan a_n}+O\left(\arctan a_n\right)$$

and by $$x=\arctan a_n \to 0$$ we reduce to

$$\frac1{e^x-1}- \frac1{x} =-\frac{e^x-1-x}{x(e^x-1)}=-\frac x{e^x-1}\frac{e^x-1-x}{x^2} \to -\frac12$$

Justification of $$(1)$$

We have that

$$\frac{\frac{1}{a_n}-\frac{1}{\arctan a_n}}{\arctan a_n}=\frac{\arctan a_n-a_n}{a_n\;\arctan^2 a_n}=\frac{-\frac13 a_n^3+O(a_n^5)}{a_n^3+O(a_n^5)}\to -\frac13$$

therefore, by definition

$$a_n^{-1} =\arctan^{-1} a_n+O\left(\arctan a_n\right)$$

• How did you get $a_n^{-1} =\arctan^{-1} a_n+O\left(\arctan a_n\right)$ (where did that $O\left(\arctan a_n\right)$ come from)? Also, did I make a mistake in proving my second question? Commented Jun 21, 2023 at 17:55
• You are right all the squared part is $O(\arctan^2 a_n)$.
– user
Commented Jun 21, 2023 at 18:00
• And what about my first question in comment above? And why are those first two equalities equivalente? I get the $\Rightarrow$ part, it follows from the fact that $\arctan(x) \geq x$ when $x \leq 0$, and all terms in my sequence are in interval $[-1, 0]$, but what about the other part? Commented Jun 21, 2023 at 18:21
• @Ranko Yes of course, it is $-\frac13$ indeed, anyway this doesn't change the result. For the other equivalence we can proceed again by the definition and note that $$\frac{a_n -\arctan a_n}{\arctan^2 a_n} \to 0$$ but at the end it is not necessary to prove (1).
– user
Commented Jun 21, 2023 at 20:21
• I was confused about the equivalence part, but this edited solution is perfectly clear to me. Thank you! Commented Jun 21, 2023 at 20:31

The context is

Given $$a_1 = -1$$ and $$a_{n+1} = e^{\arctan(a_n)}-1$$ find $$c$$ such that $$a_n \sim \frac cn$$, $$n\to+\infty$$

As a first step, negate the sequence. That is, define a new sequence

$$b_1 = 1, \quad b_{n+1} = \phi(b_n) \quad \text{ where }\quad \phi(z) := 1 - e^{-\arctan(z)}. \tag1$$

Notice the power series expansion

$$\phi(z)=z - \frac{z^2}2 - \frac{z^3}6 + \frac{7}{24}z^4 + \cdots \tag2$$

which leads to a similar kind of limiting behavior as MSE question 2861768 where $$\,\phi(z) = z-z^2.\,$$ Similar reasoning as there leads to an Ansatz

Define notation $$\,x := 1/n,\, y := \log(x).\,$$ The Ansatz is

$$b_n = f(n) := (c_1)x + (c_2+c_3y)x^2 +(c_4+c_5y+c_6y^2)x^3 + \dots. \tag3$$

The $$\,c_k\,$$ are constants to be determined by the recursion

$$f(n+1) = \phi(f(n)). \tag4$$

Now solve for these constants to get

$$f(n) = 2\,x + \frac{c + 10y}3 x^2 + \frac{(c^2+10c-70)+(20c+100)y +100y^2}{18} x^3 + \dots \tag5$$

where $$\,c\,$$ depends on the value of $$\,b_1.\,$$ In the case $$\,b_1=1\,$$ the value is $$\,c \approx 5.8291934.$$

The limiting behavior $$\,a_n \sim -2/n\,$$ comes from $$\,a_n = -b_n = -f(n) \sim -2/n .$$

I just don't get it where did $$a^{−1}_n$$ go
$$b_{n+1}^{-1} - b_n^{-1} = \phi(b_n)^{-1} - b_n^{-1} = \frac12 + \frac5{12}b_n + \cdots \tag{6}$$
Note that, similar to the MSE question I linked to, there is a nice continued logarithm expression for $$\,b_n,\,$$ namely
$$b_n = 2/(n + c_0 - c/6 + 5/3\log(n + c_1 - c/6 + 5/3\log( n + \dots ))) \tag7$$
where $$\,c_0 = 0,\, c_1 = 7/6,\, c_2 = 769/600,\, c_3 = 551587/400000,\, \dots.$$