How to find the $\lim_{n\to\infty}s_n$ when $s_1=5, s_n =\sqrt{2+s_{n-1}}$ using the Monotone Convergence Theorem?
I have the proof from my professor, but I am stuck at one step.
Proof:
Step 1: Show that it is monotonic:
Proof by induction:
Claim: $s_n > s_{n+1}, \forall n\in\mathbb{N}$
Base case: $n=1$. $s_1 = 5 > s_2=\sqrt{7}$, so it holds for $n=1$.
Assume it holds for $n=k$, now show it's true for $n=k+1$.
Assume $s_k>s_{k+1}$, then $s_{(k+1)+1} = s_{k+2} = \sqrt{2+s_{k+1}} < \sqrt{2+s_k} = s_{k+1}$ by assumption. Therefore, by the Principle of Mathematical Induction, the statement is true $\forall n\in\mathbb{N}$ and $\{S_n\}$ is decreasing (monotonic).
Step 2: Find the limit:
It is clear $0\leq s_n\leq s_1 = 5\forall n$, so $\{S_n\}$ is bounded.
Therefore, since $\{S_n\}$ is monotone and bounded, by the Monotone Convergence Theorem, $\{S_n\}$ converges.
Let $\lim_{n\to\infty} s_n = L$:
\begin{align}
\lim_{n\to\infty}\sqrt{2+s_{n-1}} &= L\\
\sqrt{2 + \lim_{n\to\infty} s_{n-1}} &= L\\
\sqrt{2+L} &= L *\text{This is where I get stuck…}\\
2+L &= L^2\\
0 &= L^2 – L – 2\\
&= (L-2)(L+1)\\
&\implies L=-1, 2
\end{align}
But $L\ne-1$ since $s_n\geq 0\forall n$. Therefore, $\lim_{n\to\infty} s_n = 2$.
I get stuck because I am unsure why we can say $\lim_{n\to\infty} s_{n-1} = L$. We said that $\lim_{n\to\infty} s_{n} = L$, but nothing about the limit of $s_{n-1}$.
Best Answer
Write down the definition of a limit to show that if $(a_n)$ is any converging sequence, then $(a_{n-1})$ is also a converging sequence and both limits are the same.