0
$\begingroup$

As I had such a great response to my last question, I'd like to check my workings are correct for this one. I have:

$\left|\frac{1}{(m+5)^3} - \frac{1}{(n+5)^3}\right|<\epsilon$

LHS $\leq \left|\frac{1}{(m+5)^3}\right| + \left|\frac{1}{(n+5)^3}\right| < \left|\frac{1}{m^3}\right| + \left|\frac{1}{n^3}\right| \leq \left|\frac{1}{n_{0}^3}\right| + \left|\frac{1}{n_{0}^3}\right| < \epsilon$

Therefore, $\epsilon > \frac{2}{n_{0}^3}$

Which leads to $n_{0} > \sqrt[3]{\frac{2}{\epsilon}}$

All comments appreciated. Thanks.

  • 5
    Maybe you can use that this sequence is convergent and hence Cauchy ? But in any case, your solution works - but in only works because the sequence converges to $0$, otherwise you should apply other methods.2012-05-31
  • 0
    Thanks, good to know it works. Our lecturer has only introduced Cauchy sequences which converge to $0$ so far so I think this is the method to use in the exam (I hope!).2012-05-31
  • 0
    To elaborate a little more on Ilya's comment. You don't even really use the difference. You show that both summands seperately go to zero. So you just show that it is a zero sequence and to talk about a Cauchy sequencen seems a bit artificial.2012-05-31
  • 3
    The *computations* are right, the writeup not so good, the logic of the argument is not given. Should start by saying that we are given an $\epsilon>0$, and want to find an $n_0$ such that $\dots$. Then forget about $\epsilon$ for a while, and (say) show like you did that if $m$, $n$ are $\gt w$ then $|a_m-a_n| \lt \frac{2}{w^3}$. And *finally* conclude that if $n_0>\sqrt[3]{2/\epsilon}$ then $\dots$.2012-05-31
  • 0
    Apologies - I missed out the 'blurb' surrounding the working. I'll remember in future.2012-05-31
  • 0
    @Sarah_24 Also, note that if $(a_n)$ is a Cauchy sequence that converges to $a$ then the sequence $(b_n)$, where $b_n=a_n-a$ is a Cauchy sequence converging to $0$.2012-05-31

1 Answers 1