I probably didn't use proper terminology. But you know, 1 is always 1 in the real numbers, but on the infinitesimal scale, 1 doesn't necessarily equal 1 exactly. Thus 1^infinity is an indeterminate form.
Well really that's a nice simple proof, the only problem is that it's already assuming it converges. And you know it converges because you know it's geometric, which means you already know its sum is .9/(1-.1)=1. So yeah, it *is* making things overly complicated, really. But then again I suppose this proof is aimed at those who don't understand the concept of convergence and don't know about geometric series. :P
But once again, you're relying on what you already know about geometric series, which means, once again, you can skip all that and go straight to the .9 repeating =1.
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Date: 2004-04-04 06:28 pm (UTC)no subject
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Date: 2004-04-04 09:28 pm (UTC)1/3 = 0.33333....
2/3 = 0.66666....
1/3 + 2/3 = 3/3 = 1
0.99999... = 0.33333... + 0.66666... = 1
?
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