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Let $a_0=3$ and $a_{n+1}=\sqrt{a_n+7}$ if $n>0$

Prove: $3<a_n<4$

At first I was quite surprised it's actually true for the base cases:

$n=0$, $a_1=\sqrt{3+7}=\sqrt{10}$

$n=1$, $a_2=\sqrt{\sqrt{10}+7}$

I'm not sure here how to even construct the induction hypothesis, never had an exercise where I needed to prove that some variable is in some range.

*Taken out of a book called "A Walkthrough Combinatorics" and unfortunately, there's no solution for this exercise.

I'm not a student yet so I don't know any university level math.

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    Hint: if $3 \lt a \lt 4$ then $\sqrt{10} \lt \sqrt{a+7} \lt \sqrt{11}$. – dxiv Aug 14 '18 at 19:48
  • This class of problems is so well covered on MSE https://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq and https://math.stackexchange.com/questions/11945/limit-of-the-nested-radical-sqrt7-sqrt7-sqrt7-cdots?noredirect=1&lq=1 – rtybase Aug 14 '18 at 21:06

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Since $a_0=3$, $a_0+7=10$ and so $9<a_0+7<16$. Therefore, $3<a_1<4$.

Now, let $n\in\mathbb N$ and suppose that $3<a_n<4$. Then $10<a_n+7<11$ and therefore $9<a_n+7<16$. So, $3<a_{n+1}<4$.

José Carlos Santos
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