Let $a_1=1$, for each $n\geq 1$ let $a_{n+1}=\sqrt{3+2a_n}$. Prove with induction that for all $n\in\mathbb{N}$, we have $a_{n}\leq a_{n+1}\leq 3.$

Asked Mar 04 '22 at 20:50

Active Mar 04 '22 at 20:50

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Let $a_1=1$ and for each $n\geq 1$ let $a_{n+1}=\sqrt{3+2a_n}$. Prove with induction that for every $n\in\mathbb{N}$, we have $$a_{n}\leq a_{n+1}\leq 3.$$

For the base case $n=1$, I already know it works. Assume $n=k$ works: $a_k\leq a_{k+1}\leq 3$. However, I'm having trouble proving that $a_{k+1}\leq a_{k+2}\leq 3.$

asked Mar 04 '22 at 20:50

MathMagician

1

Does this answer your question? Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$. It works the same way for your question. – Dietrich Burde Mar 04 '22 at 20:54
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Also this post. – Dietrich Burde Mar 04 '22 at 20:58

Let $a_1=1$, for each $n\geq 1$ let $a_{n+1}=\sqrt{3+2a_n}$. Prove with induction that for all $n\in\mathbb{N}$, we have $a_{n}\leq a_{n+1}\leq 3.$

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