|
From: Joshua Cranmer on 9 Jun 2010 23:28
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
tried it. -- Donald E. Knuth
| |