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# Dowód na zbieżności ciągu

Transkrypcja filmu video (w języku angielskim)

I made a claim that for
this sequence-- and this was in a previous video--
that for this sequence right over here that can
be defined explicitly in this way, that the
limit of the sequence-- and so I can write this as
negative 1 to the n plus 1 over n. That's one way of defining
our sequence explicitly-- the limit of this as
n approaches infinity is equal to 0. And it seems that way. As n gets larger and
larger and larger, even though the numerator
oscillates between negative 1 and 1, it seems like
it will get smaller and smaller and smaller. But I didn't prove
it, and that's what I want to do in this video. In order to prove it, this is
going to be true if and only if for any epsilon
greater than 0, there is a capital
M greater than 0 such that if lowercase n, if our
index is greater than capital M, then the nth
term in our sequence is going to be within epsilon of
our limit, within epsilon of 0. So what does that say? That says, hey, give
me-- our limit is 0. Let me do this in a new color. So our limit right
over here is 0. That's our limit. So our limit right
over here is-- we're saying the sequence
is converging to 0. What we're saying is, give
us an epsilon around 0. So let's say that this right
over here is 0 plus epsilon. That is 0 plus epsilon. The way I've drawn it here
looks like epsilon would be 0.5. This would be 0 minus epsilon. Let me make it a
little bit neater. So this would be
zero minus epsilon. So this is negative epsilon, 0
minus epsilon, 0 plus epsilon. Our limit in this case, or
our claim of a limit, is 0. Now, this is saying
for any epsilon, we need to find an M such
that if n is greater than M, the distance between our
sequence and our limit is going to be
less than epsilon. So if the distance between
our sequence and our limit is less than epsilon,
that means that the value of our sequence for
a given n is going to be within these two bounds. It's got to be in this
range right over here that I'm shading
above a certain n. So if I pick an n
right over here, it looks like anything
larger than that is going to be the case that
we're going to be within those bounds. But how do we prove it? Well, let's just
think about what needs to happen for
this to be true. So what needs to be happen
for a sub n minus 0, the absolute value
of a sub n minus 0, what needs to be true for
this to be less than epsilon? Well, this is
another way of saying that the absolute value of a sub
n has to be less than epsilon. And a sub n is just this
business right here, so it's another way of saying
that the absolute value of negative 1 to
the n plus 1 over n has to be less than epsilon,
which is another way of saying, because this negative
1 to the n plus 1, this numerator just swaps
us between a negative and a positive
version of 1 over n. But if you take the
absolute value of it, this is always just
going to be positive. So this is the same
thing as 1 over n, as the absolute value of 1 over
n has to be less than epsilon. Now, n is always
going to be positive. n starts at 1 and
goes to infinity. So this value is always
going to be positive. So this is saying the
same thing that 1 over n has to be less than epsilon
in order for this stuff to be true. And now we can take the
reciprocal of both sides. And if you take the reciprocal
of both sides of an inequality, you would have that
n-- if you take the reciprocal of both
sides of an inequality, you swap the inequality. So for this to be true, n has to
be greater than 1 over epsilon. And we essentially
have proven it now. So now we've said, look, for
this particular sequence, you give me any
epsilon, and I'm going to set M to be 1 over epsilon. Because if n is greater than
M, which is 1 over epsilon, then we know that this right
over here is going to be true. That is going to be true. So the limit does
definitely exist. And so over here, for
this particular epsilon, it looks like we've picked
0.5 or 1/2 as our epsilon. So as long as n is greater
than 1 over 1/2, which is 2, so in this case we could
say, look, you gave me 1/2. My M is going to be a
function of epsilon. It's going to be defined
for any epsilon you give me greater than 0. So here, 1 over 1/2
is right over here. I'm going to make my
M right over here. And you see it is
indeed the case that my sequence
is within the range as we passed for any
n greater than 2. So for n is equal to
3 it's in the range. For n is equal to 4
it's in the range. For n equals 5-- and it
keeps going and going. And we're not just
taking our word for it. We've proven it right over here. So we've made the proof. You give me any
other any epsilon, I said M is equal to
1 over that thing. And so for n greater than
that, this is going to be true. So this is definitely the case. This sequence converges to 0.