QUESTION: Consider the sequence

$a_1 = 24^{1/3}$ $a_{n+1} = (a_n + 24)^{1/3},n ≥ 1.$

Then what is the integer part of $a_{100}$ ?

MY APPROACH: I tried this one really hard but couldn't get the trick.. I used log, but that doesn't really help and the problem becomes more and more complex, so I am avoiding a confusing solution here..

Then I tried by defining a function say $$f(x)=(x+24)^\frac{1}3$$ Therefore by computing the derivative of $f$ we find that the rate at which the function increases, decreases with increase in x. Which also is quite clear from intuition. But I could not apply the result to solve the problem.

Can we form a recursive series for it? Any help will be much appreciated. Thank you so much.

  • 3
    $\begingroup$ Can you show the sequence is increasing, and its elements are between $2$ and $3$? $\endgroup$ Jun 5 '20 at 19:21
  • 1
    $\begingroup$ It looks like this sequence converges. Maybe you can show what it converges toward. $\endgroup$
    – Carser
    Jun 5 '20 at 19:22
  • $\begingroup$ @J. W. Tanner , I get that the sequence is increasing but how do I prove that the elements are between $2$ and $3$ ? $\endgroup$ Jun 5 '20 at 19:28
  • 1
    $\begingroup$ If it's increasing and starts with $24^{1/3}$, it has to stay more than $2=8^{1/3}$, and if a term is less than $3$ then the subsequent term is less than $(3+24)^{1/3}=3$ $\endgroup$ Jun 5 '20 at 19:34
  • $\begingroup$ Exactly! Thank you so much @J. W. Tanner $\endgroup$ Jun 5 '20 at 19:38

Hint: Prove by induction that $2 < a_n < 3$ for all $n$.

  • $\begingroup$ You can also prove that the sequence is increasing and converges to $3$, but that is irrelevant for the question. $\endgroup$
    – lhf
    Jun 5 '20 at 19:28
  • $\begingroup$ Thank you so much for the hint.. I tend to forget sometimes that there is a method called induction, when nothing can be done 😅.. so the answer to this question is $2$ right? $\endgroup$ Jun 5 '20 at 19:30
  • 1
    $\begingroup$ @StrangerForever, yes, exactly $\endgroup$
    – lhf
    Jun 5 '20 at 19:31
  • $\begingroup$ I don't know how to prove that..... Although it's irrelevant (the induction is much easier), will you tell me how to do that? learning never hurts :) $\endgroup$ Jun 5 '20 at 19:32
  • $\begingroup$ @StrangerForever, prove by induction that the sequence is increasing. Use the monotone sequence theorem to prove that it converges: en.wikipedia.org/wiki/… .Find its limit. $\endgroup$
    – lhf
    Jun 5 '20 at 19:34

If $x<3$, then $f(x)<(3+24)^{1/3}=3$. If $x\geq 0$, then $f(x)\geq(0+24)^{1/3}>2$.

So what can you say about $a_{100}=f^{100}(0)$?

  • $\begingroup$ By $f^{100}(0)$ do you mean the $100^{th}$ derivetive of $f$ ? $\endgroup$ Jun 5 '20 at 19:34
  • $\begingroup$ @Stranger Forever, no. $f^{100}$ is apply $f$ 100 times, i.e.$f^{100}(0)= f(f(...f(0)...))$. That is what you want since $a_{n+1}=f(a_n)$. $\endgroup$ Jun 5 '20 at 19:39
  • $\begingroup$ Yeah, it will be less than $3$.. I get it.. $\endgroup$ Jun 5 '20 at 19:40

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