The sequence $(x_n)_{n \in \Bbb Z_+}$ is defined recursively by $x_1 := 1$ and $x_{n+1} := \sqrt{2x_n+3}$.
- Show that sequence $(x_n)$ is bounded above by $3$.
We will use induction to prove this. We know $x_1 = 1 \leq 3$. Now lets assume $x_n \leq 3$, so we must prove $x_{n+1} \leq 3$. So for $k = n+1$
\begin{equation*}
\begin{split}
x_k\leq 3 \Longleftrightarrow 2x_k+3\leq 9 \Longleftrightarrow \sqrt{2x_k+3}\leq 3,
\end{split}
\end{equation*}
or $x_{n+1}\leq 3$ and we are done.
- Find where the quadratic polynomial $x^2-2x-3$ is negative and use this fact to show that the sequence $(x_n)$ is strictly increasing.
Not sure on this one but I know this is negative when $-1<x<2$.
- Explain why the sequence $(x_n)$ converges.
Since $(x_n)$ is (strictly) increasing and bounded above, it converges by the monotone convergence theorem.
- By taking the limit on both sides of the equation $x_{n+1}^2 = 2x_n+3$, calculate the limit $\ell = \lim_{n\to\infty} x_n$.
If we write $\ell := \lim_{n\to\infty} x_n$ then we can so some algebra to compute $\ell$.
We have
\begin{equation*}
\begin{split}
\ell = \lim_{n\to\infty} x_{n+1} &= \lim_{n\to\infty} \sqrt{2x_n+3} \\
&= \sqrt{2\lim_{n\to\infty} x_n+3} \\
&= \sqrt{2\ell+3}.
\end{split}
\end{equation*}
Thus, $\ell = \sqrt{2\ell+3}$. Thus, $\ell^2 = 2\ell+3 \Longleftrightarrow \ell^2-2\ell-3 = 0 \Longleftrightarrow (\ell-3)(\ell+2) = 0$. Solving this gives $\ell = 3$ or $\ell = -2$. Since all the terms in $(x_n)$ are positive we see that $\ell$ must be $3$.
Just looking for help with question 2. Thanks!
Best Answer
We have that $$x^2-2x-3 < 0$$ on $(-1, 3)$, which means that $$x^2 < 2x+3$$ $$x < \sqrt{2x+3}$$ So $$x_{n+1}=\sqrt{2x_n+3} > x_n$$