For this question, I'm not sure how the fact that $|a_n|<6$ effects the result of this question. I'm not really sure how to complete the proof. Here is what I have so far. Can anyone please help me out?

Consider the sequence defined by $a_{n+1} = \sqrt{2+a_n}$ if $n ≥ 1$ and $a_1 = 1$. Suppose that $|a_n| < 6$ for all n ∈ N. Does $\{a_n\}$ converge or diverge? Make sure to fully prove your claim and cite appropriate theorem(s) and hypothesis conditions.

$a_2 = \sqrt{2+1} = \sqrt{3}$

$a_3 = \sqrt{2+\sqrt{3}}$

Wts there exists an M st $a_n \le M$

Choose M = 2 since the sequence eventually apporaches 2.

Base Case: $a_1 = 1 < 2$

Induction Hypothesis: Let k ∈ N be arbitrary.

Assume $a_k \le 2$

Induction Step: $a_{k+1} \to a_k$

$a_{k+1} = \sqrt{2+a_k} < \sqrt{2+2} = 2$

Therefore by induction, the sequence is bounded above

$a_n^2 - a_{n+1}^2 = a_n^2 - \sqrt{{2+a_n}}^2 = a_n^2 - a_n - 2 = (a_n-2)(a_n+1)$

$a_n^2 - a_{n+1}^2 <0$

$a_n^2 < a_{n+1}^2$

$a_n < a_{n+1}$

Therefore by the difference test, $\{a_n\}$ is strictly increasing.

Therefore by the bounded monotone convergence theorem,$\{a_n\}$ converges.

  • 1
    $\begingroup$ Your proof is fine, so far as I see. As for $|a_n| < 6$ I guess they were just telling you to assume that is was bounded, but you've proved that. $\endgroup$ – saulspatz Apr 14 '18 at 19:18
  • $\begingroup$ It's similar to math.stackexchange.com/questions/2585780/… $\endgroup$ – rtybase Apr 14 '18 at 20:01
  • $\begingroup$ The sequence is bounded by 2, so I'm not sure what $|a_n| < 6$ means $\endgroup$ – dg123 Apr 15 '18 at 0:23

Here is a different approach that avoids induction by using Geometric Series and the Squeeze Theorem.

First, rewrite the square roots as exponents. The limit of the sequnce is given by:
$$\lim_{n\rightarrow\infty}c_{n+1}=\large2^{\sum\limits_{n=0}^{\infty}\frac{1}{2}(\frac{1}{2^n})}$$ We are only interested in whether or not the sequence converges, which is done by looking at the terms of the sequence where $n$ is large. The only difference the sequence above and the one which you have defined is that the initial term $c_{1}=\sqrt{2}$ is greater than $1$. Notice that each $c_{n}$ is larger than each $a_n$. So if we have two sequences congerving to the same limit, and $a_{n}$ is between these two sequences for each $n\in\mathbb{N}$, then $(a_{n})_{n=1}^{\infty}$ will converge to the same limit. The tables below should explain this numerically in some sense.

\begin{array}{|c|c|} \hline n & 1 & 2 & 3 & 4 \\ \hline b_{n} & 0.00000... &1.41421... & 1.74776... & 1.96157... \\ \hline \end{array} \begin{array}{|c|c|} \hline n & 1 & 2 & 3 & 4 \\ \hline a_{n} & 1.000000... & 1.73205... & 1.93185... & 1.98289... \\ \hline \end{array} \begin{array}{|c|c|} \hline n & 1 & 2 & 3 & 4 \\ \hline c_{n} & 1.41421... & 1.74776... & 1.96157... & 1.999037... \\ \hline \end{array}

Furthermore, in hopes of using the Squeeze Theorem, we intend to "squeeze" $(a_{n})_{n=1}^{\infty}$ between two other sequences (as depicted in the tables) for each $n\in\mathbb{N}$. So now for the lower sequence, we just need to set $b_{1}=0$ and then define each $b_{n+1}$ the same way as we did for the other two sequences. This will give us $b_{n}\leq a_{n}\leq c_{n}$ for all $n\in\mathbb{N}$.

Now, since the infinite series defined in the limit of $c_{n}$ can be written as: $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}=\sum\limits_{n=0}^{\infty}\frac{1}{2}\Bigl(\frac{1}{2}\Bigr)^{n}=\frac{1}{2}\Bigl(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\Bigr)$$ Then we can use the formula for Geometric Series to see that this infinite series converges to $1$: $$\sum\limits_{n=0}^{\infty}\frac{1}{2}\Bigl(\frac{1}{2}\Bigr)^{n}=\Bigl(\frac{1}{2}\Bigr)\dfrac{1}{1-(1/2)}=1$$ So $(c_{n})_{n=1}^{\infty}$ converges to $2$, and the same is true for $(b_{n})_{n=1}^{\infty}$ since each $b_{n}=c_{n-1}$ for all $n\geq 2$ Therefore by the Squeeze Theorem, the sequence $(a_{n})_{n=1}^{\infty}$ converges with $\lim\limits_{n\rightarrow\infty}a_{n+1}=\lim\limits_{n\rightarrow\infty}a_{n}=2$


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