Suppose that $-1\leq t\leq 1$. Define $t_{n}$ recursively by setting $t_{0} = 0$ and $t_{n} = t_{n-1} + \frac{1}{2}(t-t_{n-1})^{2}$. Show that $0\leq t_{n-1}\leq t_{n}\leq |t|$, for all $n\in\textbf{N}$. What is $\lim_{n\rightarrow+\infty}t_{n}$?
MY ATTEMPT
The inequality $t_{n-1} \geq 0$ is obvious, since $t_{0} \geq 0$ and each successive term is obtained from the previous one by adding a non-negative term.
Similarly, one has that $t_{n}\geq t_{n-1}$, because its difference $t_{n} - t_{n-1} = \frac{1}{2}(t-t_{n-1})^{2} \geq 0$.
On the other hand, one has that \begin{align*} 2(|t| - t_{n}) & = 2|t| - 2t_{n-1} - (t - t_{n-1})^{2}\\\\ & = 2|t| - 2t_{n-1} - |t|^{2} + 2tt_{n-1} - t^{2}_{n-1}\\\\ & = 2 + 2tt_{n-1} - (1-|t|)^{2} - (1+t_{n-1})^{2} = \ldots \end{align*}
Assuming the relation $t_{n}\leq |t|$ holds, we conclude the sequence $t_{n}$ is increasing and bounded, therefore it converges to the $\sup\{t_{n}\in\textbf{R}\mid n\in\textbf{N}\}$. According to the properties of limits, one has that \begin{align*} \lim_{n\rightarrow+\infty} t_{n} & = \lim_{n\rightarrow+\infty}(t_{n-1} + \frac{1}{2}(t - t_{n-1})^{2}) \Longrightarrow\\\\ L & = L + \frac{1}{2}(t - L)^{2} \Longrightarrow L = t \end{align*}
Can someone help me finish the proof of $|t| \geq t_{n}$ as well as double-check if I am reasoning right?
For the upper bound you can use induction:
$n=0$ is clear as $t_0 = 0 \leq |t|$.
$n=1$ is also clear since $t_1 = t^2/2 < t^2 <|t|$
Assume that, this inequality holds for $n=k$, that is $t_k < |t|. \quad$ ($1$)
We want to show $t_{k+1} < |t|$. Inspiring by what you did:
$2(|t|-t_{k+1}) = 2|t| - 2t_k -(t-t_k)^2 = 2(|t|-t_k) - (t-t_k)^2 > 2(|t|-t_k) > 0$. So we show
$|t|-t_{k+1} > 0$.
Note that we use $(1)$ in the last two inequality. Hence, by induction, $t_n < |t|$ holds for all $n \in \mathbb{N}$.
And the rest are very good reasoning to get the limit.