$a_{n+1}=\ln (1+ \arctan(a_n))$, convergence of sequence, asymptotic approximation with $cn^\alpha$, and convergence of $\sum_{n=1}^{+\infty}a_nx^n$

Question

Let $\{a_n\}_{n\in\mathbb{N}}$ be a real sequence and $a_1=1$, $a_{n+1}=\ln (1+ \arctan(a_n))$ for $n\geq 1$.

1. Prove that $\{a_n\}_{n\in\mathbb{N}}$ is convergent and find its limit.
2. Find real constants $c$ and $\alpha$ such that $a_n\sim cn^\alpha$ when $n\rightarrow +\infty$.
3. 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?
4. 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.

Answer 1

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

$$a_n=2n^{-1}+\mathcal O\left(\frac{\ln(n)}{n^2}\right)$$

1. The limit is then given by $0$.

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

3. 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.

4. It converges on $[-1,1)$ with conditional convergence at $-1$ using $a_n\sim2n^{-1}$.

Answer 2

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.

Answer 3

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 $[-1,1[$