Let $(u_n)_{n\in\mathbb{N}}$ be sequence defined as follows: $$\left\{ \begin{cases} u_0\in\mathbb{R}^{+}\\\forall n\in\mathbb{N},\quad u_{n+1}=\dfrac{e^{-u_n}}{n+1}\\ \end{cases} \right\}$$
What is the nature of $\displaystyle \sum_{n\geq 0}(-1)^{n}u_n $ ?
Here's solution provided by the author of the book:
- $\forall n\in\mathbb{N},\quad u_n\geq 0$
- $\forall n\in\mathbb{N}, 0\leq u_{n+1}=\dfrac{e^{-u_n}}{n+1}\leq \dfrac{1}{n+1}$ then $u_n\underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow}0$
- $u_{n+1}=\dfrac{e^{-u_n}}{n+1}\sim \dfrac{1}{n+1}$ then $u_n\sim\dfrac{1}{n}$
- $u_n=\dfrac{1}{n}\left(1+\dfrac{1}{n} \right)^{-1}e^{-\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)}=\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)$
- $(-1)^{n}u_n=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)$
I don't understand what this for :
- $u_{n+1}\sim\dfrac{1}{n+1} \implies u_n\sim\dfrac{1}{n}$ and how they get that
- $u_n=\dfrac{1}{n}\left(1+\dfrac{1}{n} \right)^{-1}e^{-\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)}=\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)$
is that because :
- $f_n\sim g_n$ and $g_n=\mathcal{O}(h_n)$ then $f_n=\mathcal{O}(h_n)$
- Could someone elaborate that please ?