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

Question

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.

Answer 1

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

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

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

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

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

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

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

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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

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

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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

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

Answer 2

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 $\,a_n = -b_n = -f(n) \sim -2/n .$

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Your question

I just don't get it where did $a^{−1}_n$ go

is answered by the power series expansion

$$ b_{n+1}^{-1} - b_n^{-1} = \phi(b_n)^{-1} - b_n^{-1} = \frac12 + \frac5{12}b_n + \cdots \tag{6} $$

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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 $\,c_0 = 0,\, c_1 = 7/6,\, c_2 = 769/600,\, c_3 = 551587/400000,\, \dots.$