2
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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 $\lim\limits_{n\rightarrow\infty}a_{n+1}=\lim\limits_{n\rightarrow\infty}a_{n}=2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.