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## Integral Calculus (2017 edition)

### Unit 10: Lesson 2

Infinite sequences# Convergent and divergent sequences

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.1 (EK)

, LIM‑7.A.2 (EK)

A sequence is "converging" if its terms approach a specific value as we progress through them to infinity. Get an intuitive sense of what that even means! Created by Sal Khan.

## Want to join the conversation?

- What does diverges or converges means here ?

Please help.

Thank you.(19 votes)- Converging means something is approaching something. Diverging means it is going away. So if a group of people are converging on a party they are coming (not necessarily from the same place) and all going to the party. Similarly, for functions if the function value is going toward a number as the x values get closer, then the function values are converging on that value.(58 votes)

- At3:05, why is it (-1)^(n+1)? Is that the same as (1)^(n-1)?

Thanks(14 votes)- (-1)^(n-1) does equal (-1)^(n+1) for every n value. Both equations oscillate every other N, every (-1)^ to an even number equal(-1)^ to a different even number. You can change the (n-1) to any odd number [for example (n+67) or (n-7)] and get the same answer.(18 votes)

- I thought 1 to the infinite power was an indeterminate form, so don't you need to use L'Hopital's rule to evaluate the limit in the video?(5 votes)
- Suppose we want to know the limit of a^b as x goes to infinity, where a and b are both functions of x. If we find that a approaches 1 and b approaches infinity, we have an indeterminate form, because we can't tell without further analysis whether the forces attracting a toward 1 (making the expression approach 1) are overpowered by the forces moving b toward infinity (making the expression approach infinity or zero, depending on whether a is slightly greater than or less than 1). So that is why we say 1^∞ is an indeterminate form. HOWEVER, if a is not some function that approaches 1, but is actually the number 1, then we no longer have an indeterminate form. The expression 1^b is always 1, no matter how large or small the exponent. We no longer have an infinitesimal increment away from 1 that can be overpowered by the increase of the exponent. So we have an indeterminate form when we have a base approaching 1 and exponent approaching infinity, but not when we have a base that EQUALS 1 and exponent approaching infinity.(18 votes)

- what is the difference between this statements:

a) The sequence of Bn has no limit.

b) The sequence of Bn diverges to positive infinity.

c) The sequence of Bn is simply divergent.

d) limBn=(symbol of infinity)

n--->(symbol of infinity)(5 votes)- I'll assume
`{B(n)}`

is a sequence of real numbers (but a sequence in an arbitrary metric space would be just as fine).`a)`

`{B(n)}`

has no limit means that there is no number`b`

such that`lim (n→∞) B(n) = b`

(this may be cast in terms of an epsilon type of definition).`b)`

That`{B(n)}`

diverges to`+∞`

means that for every real number`M`

there exists a real number`N`

such that`B(n) ≥ M`

whenever`n ≥ N`

.`c)`

A sequence is divergent if and only if it is not convergent, hence this means the same as`a)`

.`d)`

This means the same as`b)`

.(5 votes)

- What is the difference between convergent sequence and a converging series?(3 votes)
- A sequence is a set of numbers. If it is convergent, the
`value`

of each new term is approaching a number

A series is the sum of a sequence. If it is convergent, the`sum`

gets closer and closer to a final sum.(8 votes)

- is this a harmonic series?(2 votes)
- It is an alternating harmonic series. The signs change in every other term.(6 votes)

- don't all functions approach a number? like isn't infinity a number? cone somebody give me an example of a function that diverges.(0 votes)
- No. Not all functions approach a number as their input approaches infinity. One of the main things a function has to do to approach a number is to start to stabilize. Take sine or cosine. We know they will never output anything greater than 1, or less than -1, we are even able to compute them for any real number. But, we know that they will always fluctuate. They don't head to infinity, and they don't converge. If we were to investigate sin(x)/x, it would converge at 0, because the dividing by x heads to 0, and the +/- 1 can't stop it's approach.

A similar resistance to staying mostly still can be found in equations that diverge as their inputs approach infinity. What number does 2^x go to? (It diverges)

We might claim that it goes to 34359738368, a very big number equal to 2^35, but we 2^x, being fickle to the point of cruelty, will leave this value (already large) for a notably larger one as x goes to 36. And remember, infinity is MUCH larger than 36, and this function will continue to leave potential values in the dust, never looking back, indefinitely.

Also, just in case it isn't clear, infinity doesn't follow many of the properties that numbers follow. infinity + 1 = infinity, infinity^2 = infinity / 8, basically infinity can't be dealt with in algebra, (although L'Hopitol's rule lets us manage it), and shouldn't be thought of as a "number" in the common sense of the word.(2 votes)

- why do we use n+1 instead of n-1 when defining the power of (-1)? is it just semantics or common protocol? I know in other cases it would be n+1 is there any reason for it not to be n-1?(2 votes)
- It's completely arbitrary. Both 𝑛 – 1 and 𝑛 + 1 are valid here (as they are both odd when 𝑛 is even and vice versa) and there is no "convention" that prefers one over the other.(3 votes)

- What is the difference between finding the limit of a sequence and finding the limit of a function?(3 votes)
- sequence (1+(-1)^n) as n goes to infinity is divergent, but how to prove that(2 votes)
- For odd n, it's equal to 0. For even n, it's equal to 2. The sequence oscillates between these two values and therefore does not converge.(2 votes)

## Video transcript

Let's say I've got a sequence. It starts at 1, then let's
say it goes to negative 1/2. Then it goes to positive 1/3. Then it goes to negative 1/4. Then it goes to positive 1/5. And it just keeps going on
and on and on like this. And we could graph it. Let me draw our vertical axis. So I'll graph this
as our y-axis. And I'm going to graph
y is equal to a sub n. And let's make this our
horizontal axis where I'm going to plot our n's. So this right over
here is our n's. And this is, let's say this
right over here is positive 1. This right over
here is negative 1. This would be negative 1/2. This would be positive 1/2. And I'm not drawing the
vertical and horizontal axes at the same scale, just so that
we can kind of visualize this properly. But let's say this
is 1, 2, 3, 4, 5, and I could keep going
on and on and on. So we see here that
when n is equal to 1, a sub n is equal to 1. So this is right over there. So when n is equal to 1,
a sub n is equal to 1. So this is y is
equal to a sub n. Now, when n is
equal to 2, we have a sub n is equal
to negative 1/2. When n is equal to 3, a
sub n is equal to 1/3, which is right about there. When n is equal to 4, a sub
n is equal to negative 1/4, which is right about there. And then when n is
equal to 5, a sub n is equal to positive 1/5, which
is maybe right over there. And we keep going
on and on and on. So you see the points,
they kind of jump around, but they seem to be getting
closer and closer and closer to 0. Which would make us ask
a very natural question-- what happens to a sub n
as n approaches infinity? Or another way of saying
that is, what is the limit-- let me do this in a new color--
of a sub n as n approaches infinity? Well, let's think about if we
can define a sub n explicitly. So we can define this sequence
as a sub n where n starts at 1 and goes to infinity
with a sub n equaling-- what does it equal? Well, if we ignore
sign for a second, it looks like it's
just 1 over n. But then we seem like
we oscillate in signs. We start with a positive, then
a negative, positive, negative. So we could multiply this times
negative 1 to the-- let's see. If we multiply it times
negative 1 to the n, then this one would be negative
and this would be positive. But we don't want it that way. We want the first
term to be positive. So we say negative 1
to the n plus 1 power. And you can verify this works. When n is equal to 1, you have
1 times negative 1 squared, which is just 1, and it'll
work for all the rest. So we could write this
as equaling negative 1 to the n plus 1 power over n. And so asking what
the limit of a sub n as n approaches infinity
is equivalent to asking what is the limit of
negative 1 to the n plus 1 power over
n as n approaches infinity is going
to be equal to? Remember, a sub n, this
is just a function of n. It's a function where we're
limited right over here to positive integers
as our domain. But this is still just a
limit as something approaches infinity. And I haven't rigorously
defined it yet, but you can conceptualize
what's going on here. As n approaches
infinity, the numerator is going to oscillate between
positive and negative 1, but this denominator
is just going to get bigger and bigger
and bigger and bigger. So we're going to get really,
really, really, really small numbers. And so this thing right over
here is going to approach 0. Now, I have not proved
this to you yet. I'm just claiming
that this is true. But if this is true-- so
let me write this down. If true, if the limit of a sub
n as n approaches infinity is 0, then we can say that a
sub n converges to 0. That's another way of
saying this right over here. If it didn't, if the
limit as n approaches infinity didn't go to
some value right here-- and I haven't rigorously
defined how we define a limit-- but if this was not true, if
we could not set some limit-- it doesn't have
to be equal to 0. As long as it-- if this was
not equal to some number, then we would say that
a sub n diverges.