2 metrics on a set that induce the same topology, but a sequence is Cauchy for 1 and not the other

I need to give an example of two metrics on a set that induce the same topology, but where a sequence is Cauchy relative to one of the metrics and not the other.

Any help would be appreciated! Thanks!

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 Please make the bodies of your posts self-contained, not relying on the title for content. – Arturo Magidin Apr 6 '11 at 0:23 Closely related: math.stackexchange.com/questions/7578/… – Arturo Magidin Apr 6 '11 at 0:25

Hint: What is the topology induced on $\left.\left\{\frac{1}{n}\;\right|\; n\in\mathbb{Z}, n\gt 0\right\}$ by the standard metric?
Alternative example. Take $\mathbb{R}$ with the usual metric, and the metric $$d(x,y) = \Bigl|\arctan(x) - \arctan(y)\Bigr|.$$ Then consider the sequence $a_n = n$.
@caligurl11: "Cauchy" is not a synonym for "convergent." In any metric space, convergent implies Cauchy, but the converse is only true when the space is complete. The sequence $(1/n)$ is Cauchy in that space under the usual metric, but because the space is not complete, being Cauchy does not necessarily imply convergent. – Arturo Magidin Apr 6 '11 at 3:34
@caligirl11: Set $\epsilon=1/2$. Can you find an $N\gt 0$ such that for all $m,n\geq N$, $|a_n - a_m|\lt 1/2$? – Arturo Magidin Apr 6 '11 at 15:42