rterte wrote: Given $a\in\mathbb{R}$ and a sequence $(u_n)$ defined by \[ \begin{cases} u_1=a\\ u_n=\frac{1}{2}+\sqrt{\frac{2n+3}{n+1}u_n+\frac{1}{4}}\quad\forall n\in\mathbb{N}^* \end{cases} \] Are you sure in second condition? Maybe $u_{n+1} $, not $u_n$

Fixed. Thanks

Any Suggestions about this One? Edited: Solution Hope best for the new part b.

Attachments:

Part b.pdf (340kb)

I think we just need to solve part b then we can get the result for part a Here is my solution for this problem Solution Since: $(u_n)$ exists then $u_2$ exists So: $\dfrac{5u_1}{2} + \dfrac{1}{4} \ge 0$ or $u_1 = a \ge \dfrac{- 1}{10}$ We will prove that if $a \ge \dfrac{- 1}{10}$ then $(u_n)$ has limit If there exists $n_0 \in \mathbb{N}$ which satisfies $u_{n_0} > u_{n_0 + 1}$ So: $u_{n_0 + 1} = \dfrac{1}{2} + \sqrt{\dfrac{2n_0 + 3}{n_0 + 1} u_{n_0} + \frac{1}{4}} > \dfrac{1}{2} + \sqrt{\dfrac{2n_0 + 5}{n_0 + 2} u_{n_0 + 1} + \frac{1}{4}} = u_{n_0 + 2}$ or $u_n$ is decreasing from $n_0$ But: $u_n \ge \dfrac{1}{2}, \forall n \in \mathbb{N}$ then: $u_n$ has finite limit Let $\lim u_n = L \left(L \ge \dfrac{1}{2} \right)$ We have: $L = \dfrac{1}{2} + \sqrt{2L + \dfrac{1}{4}}$ or $L = 3$ If $u_n$ is increasing, $\forall n \in \mathbb{N}$ then: $u_n < u_{n + 1} = \dfrac{1}{2} + \sqrt{\dfrac{2n + 3}{n + 1} u_{n} + \frac{1}{4}} < \dfrac{1}{2} + \sqrt{3u_{n} + \frac{1}{4}}$ Hence: $u_n < 4, \forall n \in \mathbb{N}$ So: $(u_n)$ has finite limit Let $\lim u_n = L \left(L \ge \dfrac{1}{2} \right)$ then: $L = \dfrac{1}{2} + \sqrt{2L + \dfrac{1}{4}}$ or $L = 3$ Therefore: with $a \ge \dfrac{- 1}{10}$ then $(u_n)$ exists and has finite limit and $\lim u_n = 3$

Note that $u_{n+1}$ is the larger root of the equation $\frac{1}{2}x^2-\frac{1}{2}x-\frac{2n+3}{2n+2}u_n=0$ so one can use some graphic arguments to conclude that $u_n$ is decreasing and hence convergent putting the limit $L$ in the initial characterization and noting that the terms of this sequence are greater than $\frac{1}{2}$ one can check that the limit is $3$.