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.

- ParthKohli

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- ParthKohli

So, the sequence is \(\{2,4,8,16\cdots \}\).

- ParthKohli

Am I correct?

- ParthKohli

@agentx5 I need a genius for this.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Powers of two, while n is greater than or equal to 1. Yep! You're good

- anonymous

Infinite sequence in fact, right? (i.e.: what's the limit as n goes to infinity for 2\(^n\) ? ) ;-)

- ParthKohli

Ah! Great stuff!
\(\{ 1,2,3,4,5,6,7\} = \{n \} _{7 \ge n \ge 1}\)

- ParthKohli

It never reaches anywhere, so it's infinity.
That was an oral question :P

- anonymous

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

- ParthKohli

|dw:1342971644614:dw|

- ParthKohli

But what if \(n\) is negative?

- anonymous

n isn't negative. or zero

- anonymous

It starts at 1.

- ParthKohli

Oh, okay.

- ParthKohli

What if we have negatives included?

- anonymous

\(\{n \} _{7 \ge n \ge 1}\) <--- this however has an upper boundary

- anonymous

2^7 = 128

- anonymous

so...
{2,4,8,16,32,64,128}

- ParthKohli

The limit is 0 if we have \(\{ 2^n\}_{n \le 0}\) right?

- ParthKohli

I believe that the sequence \(\{2,4,8,16,32,64,128 \}\) doesn't have a limit.

- anonymous

Well for one thing you've got the wrong graph here, what happens when you raise to a negative power?

- ParthKohli

Exactly, we have it getting closer to 0.

- anonymous

|dw:1342971841230:dw|

- ParthKohli

Starts at 1 and keeps getting closer to 0.

- ParthKohli

Hmm.

- ParthKohli

Oops. I posted the graph for \(x^2\).

- ParthKohli

\[\lim _{x \to -\infty} 2^x = 0\]
Good enough?

- anonymous

Yep sure does, as you approach negative infinity
\[\huge \lim_{x \rightarrow -\infty} 2^x = 2^{-\infty} = (\frac{1}{2^{\infty}}) = \frac{1}{\infty} = 0\]

- anonymous

See how the Algebra keeps coming back?

- ParthKohli

It does =)

- ParthKohli

I believe that I'd owe you a lot when I get into MIT :)

- anonymous

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 ^_^

- ParthKohli

Heh. You're a nice guy!

- ParthKohli

And nice guys don't finish last ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.