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.

  • 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

1

Expanded comment by André Nicolas:

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 $\left|\frac{1}{(m+5)^3} - \frac{1}{(n+5)^3}\right|<\epsilon \quad \text{whenever } \ m,n\ge n_0 \tag1$ Then forget about $\epsilon$ for a while, and show like you did that the left hand side is less than $2/n_0^3$. And finally conclude that by choosing $n_0>(2/\epsilon)^{1/3}$, (1) becomes true.