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The sequence $(a_n)_{n \in \mathbb N}$ is defined recursively as follows: $$ \left\{ \begin{split} a_1=&1 \\ a_{n+1}=&\sqrt{a_n+2} \end{split} \right. $$ a) Prove the sequence $(a_n)_{n \in \mathbb N}$ is increasing (using induction)
b) Show that $0 < a_n < 2$ for all $n \in \mathbb{N}$ (using induction).

For a), I know how to prove it is true for the base case. And I know that next you are meant to assume if its true for that then its true for n+1 however I don't understand how to actually do this second step.

For b) I'm guessing similarly first you'd use the base case to show that it's always greater than 0 since the base case is greater than 0? But after that I'm not sure what you'd do next.

Thank you.

emonHR
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fraser92
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1 Answers1

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For $a$, you can easily see that the base case is true, that is $a_1 = 1 < a_2 = \sqrt{1+2}$. Assume that $a_n < a_{n+1}$, then $a_n + 2 < a_{n+1} + 2$ then $\sqrt{a_n + 2} < \sqrt{a_{n+1} + 2}$, then $a_{n+1} < a_{n+2}$.

For $b$, you can easily see that the base case is true, that is $0 < a_1 = 1 <2$. Assume $0 < a_n < 2$, then $2 < a_n + 2 < 4$. Root both sides, you get $\sqrt{2} < \sqrt{a_n+2} < 2$, which means that $0 < \sqrt{2} < a_{n+1} < 2$.

Ahmad Bazzi
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