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Assume that the sequence ${a_n}$ is defined recursively by $a_{n+1} = \sqrt{3a_n + 1}$ for all $n \in \mathbb N$, with $a_1 = 1$. Use mathematical induction to prove that $a_n \leq a_{n+1}$ for all $n \in \mathbb N$.

I've gotten most of the way, but I need help with the last bit. I've proven the base case, and gotten as far as:

Assume $P(k)$ is true. That is, $a_k \leq a_{k+1}$ for any $k \in \mathbb N$. Prove $P(k+1)$. That is, $a_{k+1} \leq a_{k+1+1}$.

And now I'm stuck.

MathMagician
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Ben Davis
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  • Try substituting in the definition of $a_n$ into both your "assume" and "to prove" statements. – mweiss May 04 '15 at 00:34
  • Okay I think I have got it but i just want to be sure im doing it right. For the "to prove" statement, I did sqrt(3a(n)+1) <= sqrt(3*sqrt(3a(n)+1)+1). Then i squared both sides and substituted 1 for a(n) and got the desired inequality. Is this correct? – Ben Davis May 04 '15 at 00:45
  • Why would you substitute 1 for $a_n$? $a_n=1$ only for $n=0$. – mweiss May 04 '15 at 02:17
  • Okay well is the rest of it correct then? Just keeping a(n) arbitrary and solving the inequality? – Ben Davis May 04 '15 at 02:29
  • Depends what you mean by "solving the inequality". It would help if you post your work. But you need to be careful that you don't accidentally assume the conclusion that you are trying to prove. It sounds like you have managed to derive the assumption $P(k)$ from the conclusion $P(k+1)$. If so, then you need to check that the steps of that derivation are all reversible. If they are, then you have a proof that $P(k) \implies P(k+1)$. – mweiss May 04 '15 at 02:31
  • Okay. Here we go (I apologize for my lack of formatting; I'm new to this). so a(n+1) = sqrt(3a(n) + 1). Does this mean that a(n+2) = sqrt(3a(n+1) + 1) = sqrt(3 * sqrt(3(an) + 1) + 1)

    and then to reduce you square both sides to get: 3a(n) + 1 <= 3*sqrt(3a(n) + 1) + 1?

     – Ben Davis May 04 '15 at 02:33
  • Try it this way: Start by writing down $a_{k} < a_{k+1}$. That's your induction hypothesis; you are allowed to assume it's true. Now can you find some way to manipulate that inequality until it says $\sqrt{3a_k +1} < \sqrt{3a_{k+1} + 1}$? If you can manage that, you are just a step away from the conclusion you want. – mweiss May 04 '15 at 02:38
  • Woah. That makes so much sense now. Thanks a bunch for your help! – Ben Davis May 04 '15 at 02:43

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$a_{n+1} - a_n = \sqrt{3a_n+1} - \sqrt{3a_{n-1}+1} = \dfrac{3(a_n-a_{n-1})}{\sqrt{3a_n+1}+\sqrt{3a_{n-1}+1}} \geq 0$ since by the inductive step: $a_n \geq a_{n-1}$, and the proof is completed.

DeepSea
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Any time you have $a_{n+1}=f(a_n)$ where $f$ is an increasing function, the resulting sequence is monotone. Indeed, whichever inequality you have between $a_1$ and $a_2$ will propagate to subsequent terms:

$$a_{k+1}\ge a_k \implies f(a_{k+1})\ge f(a_k)$$ or $$a_{k+1}\le a_k \implies f(a_{k+1})\le f(a_k)$$

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Assuming $a_n\geq a_{n-1}$ we have $3a_n\geq 3a_{n-1}$ and also $3a_n+1\geq 3a_{n-1}+1$. Now, since square root is a monotonic function, we get $\sqrt{3a_n+1}\geq\sqrt{3a_{n-1}+1}$. Eventually: $$a_{n+1}=\sqrt{3a_n+1}\geq\sqrt{3a_{n-1}+1}=a_{n} $$

Chris
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