– [Voiceover] Let’s think

about the infinite series, so we’re going to go from

n equals one to infinity, of one over two to the n plus n. And what I want to do

is see if we can prove whether this thing converges or diverges. And as you can imagine based

on the context of where this video shows up on

Khan Academy that maybe we will do it using the comparison test. At any point if you feel

like you can kinda take this to the finish line, feel free to pause the video and do so. So in order to kind of

figure out or get a sense for this series right over

here, it never hurts to kind of expand it out a

little bit, so let’s do that. So this would be equal

to when n equals one this is gonna be one over

two to the one plus one, so it’s gonna be one over two plus one, it’s gonna be one third

plus that’s n equals one. When n equals two it’s gonna

be one over two squared, which is four plus two plus one over six. Plus let’s see we go

to three, n equals one, n equals two, n equals

three is gonna be one over two to the third, which

is eight plus three is 11. So one over 11, maybe

I’ll do one more term. Two the fourth power is going

to be 16 plus four is 20. Plus one over 20 and obviously we just keep going on and on and on. So it looks, it feels like

this thing could converge. All of our terms are

positive, but they are getting smaller and smaller quite fast. And if we really look at

the behavior of the terms as n gets larger and

larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this kind of behaves

like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so

the infinite series from n equals one to infinity

of one over two to the n, and so when n equals one this is going to be equal to one half. When n is equal to two this is going to be equal to one fourth. When n is equal to three

this is equal to one eighth. When n is equal to four this

is equal to one sixteenth. And we go on and on and on and on. And what’s interesting about

this is we recognize it. This is a geometric

series, so let me be clear. This thing right over here,

that is the same thing as the sum of from n

equals one to infinity of one half to the n power, just

writing it in a different way. And since the absolute

value of one half, which is just one half, so

because the absolute value of one half is less than one we know that this geometric series converges,

we know that it converges And actually we even

have formulas for finding the exact sum of or to figure

out what it converges to. And so we know this thing

converges and we see that actually these two series

combined meet all of the constraints we need for

the comparison test. So let’s go back to what we

wrote about the comparison test. So the comparison test,

we have two series, all of their terms are greater

than or equal to zero. All of these terms are

greater than or equal to zero. And then for the corresponding

terms in one series, all of them are going

to be less than or equal to the corresponding

terms in the next one. And so if we look over here,

we can consider this one, the magenta series, this

is kind of our infinite series of dealing with A

sub n and that this right over here is, well I

already did it in blue this is kind of the blue series. And notice all their

terms are nonnegative and the corresponding terms

one half is greater than one third, one fourth is

greater than one sixth, one eighth is greater than one eleventh. One over two to the n is

always going to be greater than one over two to the n plus n for the n that we care about here. And since we know that

this converges, since we know that the larger one converges. It’s a geometric series

where the common ratio the absolute value of

the common ratio is less than one, since we know

that the larger series converges therefore, the

smaller series or the one where every corresponding

term is less than the one in the blue one,

that one must also converge. So by the comparison test the series in question must also converge.

Guess what I have an exam about tomorrow ðŸ˜€ Sequences and series, nice timing

Hello Khan academy, please do a video on Riemman zeta function and series, and his conjecture.Thanks. Â Â

You rock man! thanks!!!

thank you so much

amazing

ya hocam allah senden razÄ± olsun thank you very much you've been very helpful

Nice page. Please how do you determine which one you choose as An or Bn i didn't get that part well?

Why you choose 1/2^n and not 1/n?

helpful

how would you choose your Bn?