i'm trying to work on this problem but i'm stuck and can't seem to continue.

For $n=1,2,3\ldots$
$$a_1=1, a_n=\sqrt{2+a_{n-1}}\;,\qquad n > 1$$ I then have to prove that
$$a_n < 2\qquad\text{for } n=1,2,3,\ldots$$
I first do my base case:

$$a_2=\sqrt{2+1}=\sqrt{3} $$
which is less than $2$.

I then put my goal to work towards as $$a_{k+1}<2$$
Then induction goes as follows
This is where i get stuck and i'm unsure how to continue

Can i just use the fact that since $n > 1$, then
$$\sqrt{2+a_2}=\sqrt{2+\sqrt{3}}$$ which is less than two?

  • 1
    $\begingroup$ Hey, in your last couple of statements, did you mean $n\gt1$? $\endgroup$ – AnotherJohnDoe Oct 20 '17 at 15:58
  • $\begingroup$ Hi yes, in my last statement i ment n>1 $\endgroup$ – Stan K. Oct 20 '17 at 16:07
  • $\begingroup$ Also, can't i use the fact that $a_{k+1}=\sqrt{2+a_k}$ thus $a_{k+1}=a_k$? $\endgroup$ – Stan K. Oct 20 '17 at 16:08
  • $\begingroup$ Use $a_k < 2 \implies \sqrt{2+a_k} < \sqrt{2+2}=2$. $\endgroup$ – Math Lover Oct 20 '17 at 16:21
  • $\begingroup$ Not exactly; for your inductive step, your assumption is $a_n\lt 2$, which you wish to use to prove that $\sqrt{2+a_n}\lt 2$. Note that it doesn't follow that $a_{n+1}=a_n$. $\endgroup$ – AnotherJohnDoe Oct 20 '17 at 16:23

After your base case, you have your inductive hypothesis.

Assume $ a_k < 2, \ \forall k \geq 1$

From there you want to show that if its true for k, it will be true for k + 1.

Now let $n = k + 1$, so $a_{k+1} = \sqrt{2 + a_k}$, which we know becuase thats the formula we are given.

Well, we know what $a_k$ is, because thats our inductive hypothesis. so we know $2+a_k < 4$ since $a_k < 2$. The square root of a number less than 4, is less then 2. So we $a_{k+1} < 2$

So we showed that its true for a base case, and that if its true for any number we pick, it will be true for the next number. So then by induction we showed that $a_n < 2, \ \forall n \geq 1$


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