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Ok so I realize that the last term will be sqrt(6) but I just don't know how to manipulate this expressions to make it provable by induction. I tried rewriting it using exponents but had no luck with that

user120767
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  • What have you tried so far? What would you need to show to prove by induction? Such initial work should be incorporated into your question as context. – Semiclassical Sep 29 '14 at 01:02
  • You don't need induction to prove that. Does the question require you to use induction? – Robin Goodfellow Sep 29 '14 at 01:04
  • Like I said i tried rewriting the sqrt(4) using exponents. I know that the expression needs to be manipulated somehow so I can substitute k+1. I also tried squaring both sides and am trying to find a patern here but still no luck – user120767 Sep 29 '14 at 01:05
  • @Robin Yes the question requires induction – user120767 Sep 29 '14 at 01:06
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    @user120767 Sometimes it is the case that just writing things properly will help immensely. The LHS of the inequality is short for the more correct: $a_0=2$ and $a_{n+1}=\sqrt{4+a_n}$. Now proving what is asked by induction with the proper notation, is trivial. – Git Gud Sep 29 '14 at 01:07
  • can I use a0? It says n belongs to the natural numbers. – user120767 Sep 29 '14 at 01:16
  • @user120767If in this case $0$ isn't a natural number, you can just start with $a_1$.But it's unlikely that anyone would count what I wrote as incorrect. – Git Gud Sep 29 '14 at 01:18
  • Hint: $\sqrt{a_{n+1}+4}$ is lower than $3$ even if $a_n=3$ – Joao Sep 29 '14 at 05:32

1 Answers1

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The sequence we are dealing with is recursively defined as $a_n=\sqrt{4+a_{n-1}}$. We need to prove that $a_n<3$ for all $n$.

Base case: $\sqrt{4} = 2 < 3$

Inductive step: We assume that for $n=k$ we have $a_k<3$ and then attempt to show that it follows the inequality holds for $n=k+1$ as well: $$a_k<3 \implies a_k+4 < 7 \implies a_k + 4< 9 \implies \sqrt{a_k+4} < 3 \implies a_{k+1} < 3$$

Sheheryar Zaidi
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