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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.

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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. – Ilya May 31 '12 at 15:59
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!). – Sarah_24 May 31 '12 at 16:05
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. – Simon Markett May 31 '12 at 16:06
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 $\dots$. – André Nicolas May 31 '12 at 16:18
Apologies - I missed out the 'blurb' surrounding the working. I'll remember in future. – Sarah_24 May 31 '12 at 16:27
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