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Μάθημα 17: Telescoping seriesDivergent telescoping series
Telescoping series is a series where all terms cancel out except for the first and last one. This makes such series easy to analyze. In this video we take a close look at the series 1-1+1-1+1-... Δημιουργήθηκε από τον Σαλ Καν.
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Lets say that we have the sum one minus one plus one minus one plus one
and just keeps going on and on and on like that forever and we can
write that with sigma notation. This would be the sum from n is equal to
one lower case n equals one to infinity. We have an infinite number of terms here
but see this first one, we want it to be a positive one, and then we
want to keep switching terms. So we could say that this is negative 1 to
the lowercase n minus 1 power. Let's just verify that that works. When n is equal to 1, it's negative 1 to
the 0 power, which is that. When n is equal to 2, it's 2 minus 1; it's
negative one to the first power that's equal to
that right over there. So this is a way of writing this series. Now what I wanna think about is does this
series converge to an actual finite value? Or, and this is another way of saying it,
what is the sum? Is there a finite sum that is equal to this right over here, or does this series
diverge? And the way that we can think about that
is by thinking about its partial sums let me
write that down. The partial, the partial sums of this
series. And the way we can define the partial
sums, so we'll give an index here. So capital N, so the partial sum is going
to be the sum from n equals one but not
infinity but to capital N of negative 1 to the n minus 1. So just to be clear, what this means, so
the, the partial sum with just one term is just gonna be from lowercase n
equals 1 to uppercase N equals 1. So it's just going to be this first term
right over here. It's just going to be 1. The S sub two, s sub two is going to be
equal to one minus one. It's gonna be the sum of the first two
terms. S sub three, S sub three is going to be
one minus one plus one. It's the sum of the first three terms,
which is of course equal to. Equal to, let's see this equal to one. This one over here is equal to zero. S sub 4, we could keep going, S sub four
is going to be one minus one plus one minus one which
is equal to zero again. So once again, the question is, does this
sum converge to some finite value? And I encourage you to pause this video
and think about it, given what we see about the partial
sums right over here. So in order for a series to converge, that
means that the limit, an infinite series to converge, that means
that the limit, the limit, so if you're a convergence, convergence is the same
thing, is the same thing as saying that the limit as capital, the limit as capital N approaches infinity
of our partial sums is equal to some finite. Let me just write like this, is equal to
some Finite, so Finite Value. So, what is this limit going to be? Well, let's see if we can write this. So, this is going to be, let's see s sub n, if we want to write this in general
terms. We already see if s, if capital N is odd,
it's equal to 1. If capital N is even, it's equal to 0. So, we can write, lets write this down so
s sub n I could write it like this is going to be one if n odd it's equal to zero if n even. So what's the limit as s sub n approaches
infinity so what's the limit. What's the limit, as N approaches infinity
of S sub N. Well, this limit doesn't exist. It keeps oscillating between these points. You give me, you, you go one more, it goes
from 1 to 0. You give me one more, it goes from 0 to 1. So it actually is not approaching a finite
value. So this right over here does not exist. It's tempting, because it's bounded. It's only, it keeps oscillating between 1
and 0. But it does not go to one particular value
as n approaches infinity. So here we would say that our series s
diverges. Our series S diverges.