1
$\begingroup$

For this proof, I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?

Let $\{a_n\}$ be a sequence defined recursively by

$a_1 = \sqrt{6}$
$a_{n+1} = \sqrt{6+5a_n}, n = 1,2,3,...$

Prove that $\{a_n\}$ is strictly increasing and strictly bounded above

$a_2 = \sqrt{6+5\sqrt{6}}< \sqrt{6+5*15} = 9$

$a_3 = \sqrt{6+5a_2}$

$a_3 = \sqrt{6+5\sqrt{6+5 \sqrt{6}}}$

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

Choose $M = 9$

Base Case: $a_1 = \sqrt{6}<\sqrt{81} = 9$

Induction step: Let k in the natural numbers be arbitrary

Assume $a_k \le 9$

$a_{k+1} = \sqrt{6+5a_k} < \sqrt{6+75} = \sqrt{81} = 9$

Therefore by induction, $a_n \le 9$

$a_{n+1} = \sqrt{6+5a_n}$

$a_{n+2} = \sqrt{6+5a_{n+1}}$

$a_{n+1} \le a_{n+2}$

Therefore, the sequence is strictly increasing and strictly bounded above

$\endgroup$
13
  • $\begingroup$ I don't understand your calculation. You appear to assert that $a_2=5\sqrt 6$ but this is not true. $\endgroup$ – lulu Mar 31 '18 at 18:34
  • $\begingroup$ its root 6 plus five times root 6 $\endgroup$ – dg123 Mar 31 '18 at 18:49
  • $\begingroup$ Your formula for $a_3$ incorrectly replaces $a_2$ with $5\sqrt 6$. $\endgroup$ – lulu Mar 31 '18 at 18:50
  • $\begingroup$ Check the edit I made. $\endgroup$ – dg123 Mar 31 '18 at 18:54
  • $\begingroup$ I changed it, take a look $\endgroup$ – dg123 Mar 31 '18 at 18:56
1
$\begingroup$

It is correct globally, but I don't think you are not expressing yourself clearly about why the sequence is strictly increasing. The base case is $a_1<a_2$, which is clearly true. Andn, if $a_n<a_{n+1}$, then $6+5a_n<6+5a_{n+1}$ and therefore $\sqrt{6+5a_n}<\sqrt{6+5a_{n+1}}$. But this means that $a_{n+1}<a_{n+2}$.

$\endgroup$
8
  • $\begingroup$ Since that is true, doesn't it mean it's strictly increasing? $\endgroup$ – dg123 Mar 31 '18 at 18:50
  • $\begingroup$ @dg123 Yes, it is strictly increasing. I was commenting your proof, not the statement itself. $\endgroup$ – José Carlos Santos Mar 31 '18 at 18:52
  • $\begingroup$ So what is wrong with my proof? $\endgroup$ – dg123 Mar 31 '18 at 18:54
  • $\begingroup$ @dg123 First: you did not mention that $a_1<a_2$. Second: you didn't justify the equality $a_{n+1}\leqslant a_{n+2}$. $\endgroup$ – José Carlos Santos Mar 31 '18 at 18:56
  • 1
    $\begingroup$ @dg123 If intuition is enough for you, then you don't need proofs, right? $\endgroup$ – José Carlos Santos Mar 31 '18 at 18:59

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.