$a_{n+1}=\ln (1+ \arctan(a_n))$, convergence of sequence, asymptotic approximation with $cn^\alpha$, and convergence of $\sum_{n=1}^{+\infty}a_nx^n$ Let $\{a_n\}_{n\in\mathbb{N}}$ be a real sequence and $a_1=1$, $a_{n+1}=\ln (1+ \arctan(a_n))$ for $n\geq 1$.


*

*Prove that $\{a_n\}_{n\in\mathbb{N}}$ is convergent and find its limit.

*Find real constants $c$ and $\alpha$ such that $a_n\sim cn^\alpha$ when $n\rightarrow +\infty$.

*Does  $\sum_{n=1}^{+\infty}(-1)^n\arcsin(\frac{1}{\sqrt n})\cos{a_n}$ converge, if it does converge is it absolute or conditional?

*Find all $x \in \mathbb{R}$ such that $\sum_{n=1}^{+\infty}a_nx^n$ converges


I am able to do 1. and also I am able to show that 3. converges but not if it does so absolutely. 2. and 4. I am completely unable to do.
 A: The Taylor expansion is given by
$$\ln(1+\arctan(x))=x-\frac12x^2+\mathcal O(x^4)$$
and the reciprocal expands as
$$\frac1{\ln(1+\arctan(x))}=\frac1x+\frac12+\mathcal O(x)$$
Letting $a_n=b_n^{-1}$, we then have
$$b_{n+1}=\frac1{\ln(1+\arctan(b_n^{-1}))}=b_n+\frac12+\mathcal O(b_n^{-1})$$
from which we can deduce that
$$b_n=\frac12n+\mathcal O\left(\ln(n)\right)$$
and
$$a_n=2n^{-1}+\mathcal O\left(\frac{\ln(n)}{n^2}\right)$$


*

*The limit is then given by $0$.

*We have $a_n\sim2n^{-1}$.

*It does not converge absolutely, as $\cos(a_n)\to1$ and $\arcsin(n^{-1/2})\sim n^{-1/2}$ gives divergence by the limit comparison test.

*It converges on $[-1,1)$ with conditional convergence at $-1$ using $a_n\sim2n^{-1}$.
A: Note that for $0<u\le 1$ $$\ln (1+\tan^{-1}u)\le \tan^{-1}u\le u$$hence the sequence is decreasing and bounded below by zero, therefore it tends to some $l\ge0$. The limit must satisfy the recurrence and the only such $l$ is $0$, for which all the inequalities hold with equality. From this point, we can say that $\cos a_n\to 1$ and the mentioned summation is not absolutely convergent.
A: First by iteration
$a_n$ is stricly positive $\forall n \in \mathbb{N}$
1. We know that $ \forall x\in \mathbb{R}^+ , \ \ln(1+x)\leq x$
So beacuse $a_n$ is positive we can use the inequality above, $ \forall n\in \mathbb{N}, a_{n+1}\leq a_n $ so $a_n$ is decreasing.
Because it is minored by $a_1$ (because decreasing), it is convergent (decreasing and minored).
The limit is by defintion a fix point defined by
$x=\ln(1+\arctan(x))$ , and $0$ is the only solution (the limit). It is unique because the limit of $a_n$ is unique)
2. The general techniques is like that. Let $\alpha \in \mathbb{R}$
Calculate $\dfrac{1}{a_{n+1}^\alpha}-\dfrac{1}{a_n^\alpha}$
$$ \dfrac{1}{a_{n+1}^\alpha}-\dfrac{1}{a_n^\alpha}=\dfrac{1}{(\ln(1+\arctan(a_n))^\alpha}-\dfrac{1}{a_n^\alpha} $$
Making two asymptotic development on left terms. First on $\arctan$ at order 2. Second $\ln$ at order 1.(possible because $a_n$ tends to $0$)
You find
$$ \dfrac{1}{a_{n}^\alpha(1-\frac{a_n^2}{3}+o(a_n^2))}-\dfrac{1}{a_n^\alpha} $$
So you get
$$ \dfrac{1}{a_{n+1}^\alpha}-\dfrac{1}{a_n^\alpha}=\dfrac{1}{a_n^\alpha}(\dfrac{1}{1-\frac{a_n^1}{2}+o(a_n^2))}-1)=\dfrac{1}{a_n^\alpha}(1-\dfrac{a_n^1}{2}-1 +o(a_n^2))$$
Now choosing $\alpha=-1$ (to cancel out terms in development) and using telescopic sum  :
$$ \sum_{k=1}^n \dfrac{1}{a_{n+1}^\alpha}-\dfrac{1}{a_n^\alpha}=\dfrac{n}{2}=\dfrac{1}{a_{n+1}^\alpha}-1=\dfrac{n}{2} +o(n)$$
(Where $o$ summation theroem work because of divergence of the serie we sum$
So $a_n \sim 2n^{-1}$
Hence your values.
3
It doesn't converge absolutely because you sequence in the series $b_n$ is a $O(n^{\frac{-1}{2}})$
Conditionnaly :
$$(-1)^n\arcsin(\frac{1}{\sqrt(n)})\cos(a_n)=\frac{(-1)^n}{\sqrt(n)}-\dfrac{(-1)^nc^2}{2n\frac{5}{2}}+ o(n^\frac{5}{2}) $$
and developing terms converge with alternative series criterion.
Since $y_n$ (each developed term) respond to alternate series criterion,

*

*$|y_n|$ decreasing

*$y_n$ tends to zero.

it shows conditional convergence
4
The radius of convergence of your series is $1$ since $\alpha > -1$ comparing to Riemann series or by d'Alembert criterion.
Since $x_n\triangleq a_n(-1)^n$ respond to alternate series criterion,

*

*$|x_n|$ decreasing


*$x_n$ tends to zero.
it converges at $x=-1$
So your set is $[-1,1[$
