From: Joshua Cranmer on
On 06/09/2010 09:28 PM, Herb wrote:
> lim n->inf { ( 1 + 2 + ... + n ) / n^2 }
>
> = lim n->inf (1/n^2) + lim n->inf (2/n^2) + ... + lim n->inf (n/
> n^2)

Let me ask you... how many limits are you taking the sum of?

--
Beware of bugs in the above code; I have only proved it correct, not
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