1
$\begingroup$

This question already has an answer here:

I am trying to show that this sequence $\{a_n\}$ = (2n+1)/(3n+5) does not converge to $42$ if it is not bounded above. I have already showed that it converges to $2/3$. For this I want to use a proof by contradiction,i.e, I assume that the sequence does converge to $42$, which will lead to a contradiction to the initial assumption that is the sequence will be bounded above. Any ideas or theorems that can help me solve this question?

$\endgroup$

marked as duplicate by Ted Shifrin, Rick Decker, Vedran Šego, Nick Peterson, Rebecca J. Stones Sep 17 '13 at 0:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If you've already shown it converges to $\frac{2}{3}$, assuming it converges to $42$ should yield a contradiction very quickly... $\endgroup$ – MartianInvader Sep 16 '13 at 23:00
  • $\begingroup$ Yes, but I'm finding the solution quite tedious. Is there another efficient approach to this problem? $\endgroup$ – user87274 Sep 16 '13 at 23:01
  • 2
    $\begingroup$ You've posted the same question twice. math.stackexchange.com/questions/495784/… ... And you've made a total hash of it the second time. Please do not repost. $\endgroup$ – Ted Shifrin Sep 16 '13 at 23:06
  • $\begingroup$ Well, sorry. I didn't know if it is such a big deal. $\endgroup$ – user87274 Sep 16 '13 at 23:26
  • $\begingroup$ @mespebjidom: You might want to take a look at this question and its answers. $\endgroup$ – robjohn Sep 17 '13 at 12:40
0
$\begingroup$

We will show that there is an $\epsilon\gt 0$ such that $$\left|\frac{2n+1}{3n+5}-42\right|\gt \epsilon$$ for infinitely many $n$.

Pick $\epsilon=1$. It is obvious that for positive $n$, we have $\frac{2n+1}{3n+5}\lt 1$. It follows that for all positive $n$, we have $\left|\frac{2n+1}{3n+5}-42\right|\gt 41\gt \epsilon$.

$\endgroup$
  • $\begingroup$ So is the unboundedness part redundant? I thought that I could use proof by contradiction by showing that if the sequence converges to 42, then there's a contradiction in the initial assumption of the sequence not being bounded above. $\endgroup$ – user87274 Sep 16 '13 at 23:20
  • $\begingroup$ You had removed that part by the time I answered, which is a good thing, since it was not useful. The sequence is bounded above, but that does not show it doesn't have limit $42$. We can use the fact it is bounded above by $1$, as I did in a formal $\epsilon$-$N$ proof. With the right theorems, you can bypass all this stuff. For example, if you have the theorem that a limit, if it exists, is unique, then you can immediately conclude from your previous work that the limit is not $42$. $\endgroup$ – André Nicolas Sep 16 '13 at 23:44
  • $\begingroup$ The expression $\frac{2n+1}{3n+5}$ is less than $1$ for any positive $n$, since the top $2n+1$ is less than the bottom $3n+5$. The choice of $\epsilon=1$ has nothing to do with the size of $(2n+1)/(3n+5)$. If it confused you, let $\epsilon=1/7$. In the formal definition of $\lim a_n=a$, we say that whatever positive $\epsilon$ we pick, after a while $|a_n-a|$ is guaranteed to be $\lt \epsilon$. To prove the limit is not $42$, we pick a convenient $\epsilon$ (I used $1$, you can use $1/13$ if you like) and show that however far out we go, (Cont) $\endgroup$ – André Nicolas Sep 17 '13 at 21:51
  • $\begingroup$ (Cont) there will be some $a_n$ whose distance from $42$ is greater than $\epsilon$. In our case, that's easy, for $a_n\lt 1$, so it is always quite far from $42$. $\endgroup$ – André Nicolas Sep 17 '13 at 21:53
  • $\begingroup$ Thanks, I figured it out and actually was silly of me to ask. $\endgroup$ – user87274 Sep 17 '13 at 21:58
1
$\begingroup$

Your approach seems distinctly strange. For one thing, if the sequence converged to $42$, then it would be bounded above!

On the other hand, you have a specific sequence that you already know is converging to $\frac{2}{3}$, so assuming that it converges to something else is simply contradictory (I assume you know that limits are unique).

Let's back up several steps. Try to show that a convergent sequence is bounded above: that's logically equivalent to your title question and less convoluted. Can you do that?

$\endgroup$
  • $\begingroup$ Check out my previous post: math.stackexchange.com/questions/495784/… $\endgroup$ – user87274 Sep 16 '13 at 23:04
  • 1
    $\begingroup$ @mespebjidom: OK, it seems that much of the strangeness is coming from the question and not from you. Still my recommendation remains the same: show that a convergent sequence is bounded above. Do you know how to show this? Do you agree that it would answer the question? If so, we're good, and it's certainly a better moral to take away than something about "convergence to 42"... $\endgroup$ – Pete L. Clark Sep 16 '13 at 23:08
  • $\begingroup$ Yeah I think should've asked: "If the sequence is not bounded, then how can I prove that it does not converge to 42?" $\endgroup$ – user87274 Sep 16 '13 at 23:12
  • $\begingroup$ @mespebjidom: you could have asked that, but it's logically equivalent to ask "Show that a sequence which converges to $42$ must be bounded," which we can immediately touch up to "Show that a sequence which is convergent must be bounded". That last statement sounds like a useful fact and not a curiosity. Again, let me know if you want help on how to show that. $\endgroup$ – Pete L. Clark Sep 16 '13 at 23:31
  • $\begingroup$ There's a theorem in my book that says that "If a sequence {an} converges, then {an} is bounded both above and below.". So in a nutshell, that's all I had to apply. $\endgroup$ – user87274 Sep 18 '13 at 17:31