0
$\begingroup$

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.

$\endgroup$
5
  • 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
2
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ 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
0
$\begingroup$

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 $a_{100}=f^{100}(0)$?

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.