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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 2: Infinite sequences# Proving a sequence converges using the formal definition

Applying the formal definition of the limit of a sequence to prove that a sequence converges. Created by Sal Khan.

## Want to join the conversation?

- what is the proof for showing sin(x) is divergent(12 votes)
- A function is divergent if it fails to converge to a single number. It doesn't have to veer off to some large value to be considered divergent. The function sin(x) oscillates between 1 and -1 forever, so it never converges to a single number.(33 votes)

- I understand everything, but how to determine the value of epsilon?(11 votes)
- Well, epsilon is an arbitrary value that is chosen for the purpose of proving a limit. In this case, Sal started drawing a line for his arbitrary value, and then said it looked like he had chosen a value of 1/2 for his epsilon. He could have chosen another value, but it is a good idea not to go too far along in the sequence, or you might wonder whether the proof holds only in the tail end of the sequence or perhaps something interesting happens in the first values.

Once he "chose" this value, he then used it as he talked through the proof. When he chose 1/2, the reciprocal was 2, so that was where he drew his M.

If he had chosen an epsilon of 1/3, his value of M would be 3, and it is still true that for values of n greater than M, the values of the sequence would be within the range defined by that epsilon. By looking at his graph, you can see that the first value was outside his criteria according to the epsilon he chose, but for convergence, we really care about what happens further along in the sequence.

Having said that, if you have to do this for another problem, you can solve the inequality to find suitable values of epsilon for the particular sequence.(25 votes)

- why does the exponent n+1 disappear after you introduce the insertion of the obsolete value signs?(6 votes)
- The exponent n+1 is only making the sequence oscillate between positive and negative (i.e. the first term is positive, the second is negative, the third is positive, the fourth is negative, etc.). However, the absolute value of any number is (by definition) positive, so the exponent has no purpose anymore (because the "new sequence" is now entirely positive and does not oscillate between positive and negative).(14 votes)

- If a sequence is undefined at a certain integer value e.g ((-1)^n)/4-n^2 is clearly undefined at n=2. What are the implications of this in regard to the sequence and its convergence?(4 votes)
- If you would draw it, a clear asymptote would be seen at 2, but after that it normally converges to +/- 0. Since you care about the convergence when n goes to infinity, a tinsy number like 2 has no significance there. To give a full proper answer though, you'd have to split it into three questions and solve them separately, |f(n)| < 2, |f(n)| = 2 and |f(n)| > 2 . The middle case has another twist, since the limit is different if you approach it from the left or right.(6 votes)

- Sal begins the video by "claiming" that the limit as the alternating harmonic sequence approaches infinite terms is zero, and then proceeds to plug in L=0. But suppose, for some reason, you thought that L=1. If you then plug that in, what would break down?(5 votes)
- To show that 1 is not the limit, we would need to find an ε>0 and an M>0 such that, for all n>M, |a_n-1|>ε. This is the negation of the limit definition.

If we take ε=1/2, M=3, we just need to show that |(-1)ⁿ/n -1|>1/2 for all n>3. We can prove this by induction or just observe that the numbers within a distance 1/2 of 1 are those in the interval (1/2, 3/2), which the remainder of this sequence stays outside of.(3 votes)

- How would you prove a sequence that's limit as n-> infinity = infinity(3 votes)
- Let a_n be a sequence. To prove that limit as n-> infinity a_n = infinity, show that for every positive number M, there exists a positive integer N such that a_n > M for all n > N. In more concrete language, this means showing that for every threshold level (M), there is a position in the sequence (N) beyond which the terms of the sequence are above that threshold level forever.

Have a blessed, wonderful day!(4 votes)

- So it's true is M is positive and not true when M is negative?(1 vote)
- Sal's statement that M > 0 is superfluous. This sequence happens to start at 1, so our formula for M will produce a positive number, but we could just as easily have a sequence that begins with a negative number and then it would be acceptable for M to be negative. For example, if this sequence were shifted 10 units to the left, then our formula for M would be 1/ε - 10, and there would be no problem in the fact that some values of ε, such as 1/4, would produce a negative M.(7 votes)

- So all the proof is saying is that if you set up some boundaries for a function and the values of the function stay in that range the function will converge at some point ?(4 votes)
- In a sense, yes, in that L + ε and L - ε are your boundaries, and as the boundaries become increasingly stringent, in other words as ε approaches 0, there must be a value of n that is greater than M for which | a sub n - L | < ε. As a result, you must be able to find a function, M(ε), that defines M for any value of ε. In the video, this function was 1/ε. If there is a value of ε where a value of M cannot be found, then the sequence is not convergent.(1 vote)

- Wouldn't the series (-1)^n converge according to the definition if I pick epsilon=2, m=1?

The sequence (1,-1,1,-1,1,-,1...) is always epsilon within 0, but it obviously doesn't converge to 0(2 votes)- No, the series does not converge. Neither does the sequence.

But, remember this video was about whether the sequence converges, not the series.

Remember this test must work for ALL ε > 0, not just a convenient value. So, you can pick 0.01 for epsilon, which obviously won't work.

So, what is L? 0? 1? A simple check will show it is none of these:

Let us pick ε = 0.01

Try L = 1

[an - 1| < 0.01

an will oscillate between -1 and 1, when an is -1, the test fails.

Try L = 0, when an is 1 or -1 the test fails.

In fact, try any value of M whatsoever and the test will fail with a sufficiently low ε no matter what we pick for a potential L.

Thus, this sequence diverges.(4 votes)

- is M a the intersection of epsilon and the function? is M a value of the sequence?(2 votes)
- M is a value of n chosen for the purpose of proving that the sequence converges. In a regular proof of a limit, we choose a distance (delta) along the horizontal axis on either side of the value of x, but sequences are only valid for n equaling positive integers, so we choose M. We have to satisfy that the absolute value of ( an - 0) < epsilon for any number greater than M, We choose epsilon as any positive number. M will be the smallest possible integer larger than 1/epsilon(4 votes)

## Video transcript

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.