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Notation of a sequence: What do we mean by \(n \ge 1\) in \(\{ 2^n\}_{n \ge 1}\)? Is it all the terms of the sequence only have \(n = 1\) or more? So basically \(n\) iterates to the positive numbers.

Mathematics
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So, the sequence is \(\{2,4,8,16\cdots \}\).
Am I correct?
@agentx5 I need a genius for this.

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Other answers:

Powers of two, while n is greater than or equal to 1. Yep! You're good
Infinite sequence in fact, right? (i.e.: what's the limit as n goes to infinity for 2\(^n\) ? ) ;-)
Ah! Great stuff! \(\{ 1,2,3,4,5,6,7\} = \{n \} _{7 \ge n \ge 1}\)
It never reaches anywhere, so it's infinity. That was an oral question :P
This is how you typically see these though, as series (sums of the terms in a sequence) \(\sum_{n \ge 1}^{\infty} 2^n\) = \(\infty\) or "Diverges" as they say
|dw:1342971644614:dw|
But what if \(n\) is negative?
n isn't negative. or zero
It starts at 1.
Oh, okay.
What if we have negatives included?
\(\{n \} _{7 \ge n \ge 1}\) <--- this however has an upper boundary
2^7 = 128
so... {2,4,8,16,32,64,128}
The limit is 0 if we have \(\{ 2^n\}_{n \le 0}\) right?
I believe that the sequence \(\{2,4,8,16,32,64,128 \}\) doesn't have a limit.
Well for one thing you've got the wrong graph here, what happens when you raise to a negative power?
Exactly, we have it getting closer to 0.
|dw:1342971841230:dw|
Starts at 1 and keeps getting closer to 0.
Hmm.
Oops. I posted the graph for \(x^2\).
\[\lim _{x \to -\infty} 2^x = 0\] Good enough?
Yep sure does, as you approach negative infinity \[\huge \lim_{x \rightarrow -\infty} 2^x = 2^{-\infty} = (\frac{1}{2^{\infty}}) = \frac{1}{\infty} = 0\]
See how the Algebra keeps coming back?
It does =)
I believe that I'd owe you a lot when I get into MIT :)
I'm used to tutoring students with learning disabilities :-) That's my part-time job on campus. Maybe you can help me find a job at some point in the future, ya never know ^_^
Heh. You're a nice guy!
And nice guys don't finish last ;)

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