1
$\begingroup$

Just wanted to make sure my proof holds up, not sure if it does.

Given $\epsilon > 0, \exists N \ s.t. \forall n \geq N:$

$|x_n - \frac{1}{5}| < \epsilon < \epsilon + \frac{4}{5} $ then we know

$ x_n < \epsilon + 1$ and that $x_n$ is bounded above by $\epsilon + 1$. Then it follows that, $\exists N \ s.t. \ \forall n \geq N$

$|\frac{1}{x_n} - 5| = |\frac{5x_n - 1}{x_n}| = \frac{5|x_n \ - \frac{1}{5}|}{|x_n|} < \frac{5|x_n \ - \frac{1}{5}|}{|\epsilon + 1|} < 5\frac{\epsilon}{\epsilon + 1} < 5\epsilon$.

I revised the proof for general $\epsilon$

My revised version also proved flawed since I made a naive assumption that if $|x_n| < \epsilon + 1$ then $\frac{1}{x_n} < \frac{1}{\epsilon + 1}$ when in fact the complete opposite is true.

However, there was a simple fix to my original proof given below.

  • 0
    (a) How did you get $\frac{5|x_n - \frac{1}{5}|}{|x_n|} < \frac{5|x_n - \frac{1}{5}|}{|5|}$ ? (b) You've *fixed* $\epsilon$ all along. Remember that you need to find an $N(\epsilon)$ *for every* $\epsilon>0$. – 2017-01-26
  • 0
    You don't get to pick the $\varepsilon$ you want. Proving the convergence implies that for **any** (every) $\varepsilon>0$ that is given to you, you can find some $N\geq 0$ (which can depend on $\varepsilon$) such that... – 2017-01-26
  • 0
    Yes, yes I see I've made a fatal mistake. However, if I don't fix epsilon it seems like this logic might still hold. – 2017-01-26

1 Answers 1

0

Your proof is incorrect, as pointed in the comments. Here's a way to fix it.

Since $x_n\to\frac{1}{5}$, taking $\epsilon=1$, there exists $N_1$ for which $|x_n-\frac{1}{5}|<1$ whenever $n\geq N_1$. So $|x_n|>\frac{1}{10}$ whenever $n\geq N_1$.

Let $\epsilon>0$. Since $x_n\to\frac{1}{5}$, there exists $N_{\epsilon}$ for which $|x_n-\frac{1}{5}|<\epsilon$ whenever $n\geq N_{\epsilon}$. So $$ \left|\frac{1}{x_n}-5\right|=\frac{5|x_n-\frac{1}{5}|}{|x_n|}<50\epsilon $$ whenever $n\geq\max(N_1,N_{\epsilon})$. This is enough to conclude $\frac{1}{x_n}\to5$.

  • 0
    Yes, I see I just needed to specify the max to take care of $\epsilon = 1$ with $N_1$ and then all other $\epsilon$ with $N_\epsilon$ – 2017-01-26
  • 0
    @ddog Your revised proof is still incorrect. First, even if $x_n<\epsilon+1$ whenever $n\geq N$, you should show that $x_n\geq0$ whenever $n\geq N$ in order to deduce $|x_n|<\epsilon+1$. Second, $|x_n|<\epsilon+1$ does not imply $\frac{1}{|x_n|}<\frac{1}{\epsilon+1}$. In fact, it is the opposite inequality that you get. If you want $\frac{1}{|x_n|}<\frac{1}{\epsilon+1}$ then you should show $|x_n|>\epsilon+1$. Third, arriving at $<5\epsilon$ at the end would be sufficient but of course $5\epsilon$ is not less than $\epsilon$ as you wrote. – 2017-01-26
  • 0
    I don't follow. If $|x_n - \frac{1}{5}| < \epsilon$ and $\epsilon > 0$, then shouldn't $|x_n - \frac{1}{5}| < \epsilon + \frac{4}{5}$ also hold? If it does, then don't we know that $-\epsilon - \frac{4}{5} < x_n - \frac{1}{5} < \epsilon + \frac{4}{5}$ and thus $-\epsilon - \frac{3}{5} < x_n < \epsilon + 1$ and so $x_n < \epsilon + 1$? – 2017-01-26
  • 0
    @ddog All that you wrote in your last comment is correct. None of my remarks had anything to do with this though. – 2017-01-26
  • 0
    Sorry, I'm a tad rusty with absolute values. I think I assumed that if $x_n < \epsilon + 1$ then $|x_n| < |\epsilon + 1|$, which should be the same as $|x_n| < \epsilon + 1$. Is it that step that you are saying is not correct? – 2017-01-26
  • 0
    @ddog If $x_n=-10$ and $\epsilon=1$ then $x_n<\epsilon+1$ and yet $|x_n|>|\epsilon+1|$. You should reread my first comment on this answer. I think I made it very clear where you misstepped. – 2017-01-26
  • 0
    I see what you are saying, but also your example is not applicable since if $\epsilon = 1$ $x_n > \frac{-8}{5}$ and if that were the case than the inequality would hold: $|\frac{-7}{5}| < 1 + 1$. Nevertheless, I see what you mean about the sign should be flipped. Thank you for the help :) – 2017-01-26
  • 0
    @ddog My point was that you should have written this argument in your proof since *in general* you can't be sure that $a$x_n$ and $\epsilon$, it will hold. But you have to make this clear. You're welcome. – 2017-01-27
  • 0
    Since comments are not meant to last and may disappear, you should include in your answer what has been said there -- i.e., why the proof is currently flawed. – 2017-01-27
  • 0
    @Clement C. I added the resolution in the question, is that ok? New here not sure whats right, sorry. – 2017-01-27