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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?


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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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 $\left|\frac{2n+1}{3n+5}-42\right|\gt 41\gt \epsilon$.

  • $\begingroup$ 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

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?

  • $\begingroup$ Check out my previous post: math.stackexchange.com/questions/495784/… $\endgroup$ – user87274 Sep 16 '13 at 23:04
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    $\begingroup$ @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