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.
Best Answer
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 $a_n\sim2n^{-1}$.