Sequences

Let's talk a about sequences. Let's start with what they are good for.

Well, for example, they are good for talking about them.

Here is a sequence.

This is the index of the sequence, and it tells us which member of the sequence we are looking at.

index

One of the most important properties of a sequence is what happens to it when we look at its members that are farther and farther away.

This sequence, for instance, is approaching zero.

So much so, that we could pick any tiny number, and still, there will be a time when the sequence will get closer than that to zero.

In this case the sequence is said to "converge to" or "tend to" zero, or "the limit of the sequence is zero".

It is denoted by this:  or this:

Here comes another sequence.

This sequence converges to zero even more.

Well, these sequences usually converge to zero.

There are then these sequences.

They tend to infinity.

LIMITS OF FAMOUS SEQUENCES

There are then such sequences with radicals.

They also tend to infinity.

And here is the most exciting sequence:

If , then

nowhere

And now let's see what happens if we add up two sequences.

For example  and , then it seems logical that

But unfortunately, life is more complicated.

It is possible, that  and .

Where does their sum converge in this case?

Well, the truth is that sequence  could converge to negative infinity,

to a specific number,

or to positive infinity.

It is true for sequence  as well.

As to their sum, there could be these nine cases.

Let's see them.

If both sequences tend to negative infinity, then their sum does, too.

If one of them tends to A, and the other to negative infinity, then their sum tends to negative infinity.

If one of the sequences tends to negative infinity, and the other to positive infinity, then we simply don't know where their sum tends.

It could be negative infinity

it could be 42

and it could be positive infinity as well.

Filling the rest of the table does not provide many surprises. The lower left corner is also a question mark.

Now let's see what it looks like for the product of two sequences.

Unfortunately there will be many cases here.

Well, this is again something that we simply don't know.

The rest is not too exciting:

And then finally some straightforward cases:

Now we will tackle the worst one: division.

There will be quite a few question marks here.

The first one, right away:

But there are more.

Well, the point of these tables is to help us navigate among the various types of limits.

The question marks aren't too helpful, so we will spend some time on figuring out what can be done in those cases.

We will start with one of the most exciting cases: .

If k IS A SPECIFIC NUMBER

                                                                                              IF

Let's see what should be done with limits of this type:  

Here is a case like that:

Let's see what should be done with limits of this type:  

Here is a case like that:

The trick is to divide by .

Both the numerator and the denominator.

This way we turned  into .

The latter is easier to figure out, we know it is 2.

Let's see another one.

Let's divide this one by , too.

Let's see what we get.

The numerator tends to 4.

The denominator tends to zero.

Well, this is a problem.

The cause of the problem is that the highest degree member in the denominator is quadratic.

So we should not be surprised that if we divide by , then everything in the denominator will tend to zero.

If we don't want the denominator to tend to zero, then we have to divide by the highest degree member of the denominator.

Divide both the numerator and the denominator by the highest degree member of the denominator.

Divide both the numerator and the denominator by the member of the denominator that has the highest base.

First we transform.

Then divide.

DIVIDE BOTH THE NUMERATOR AND THE DENOMINATOR BY THE MEMBER OF THE DENOMINATOR THAT HAS THE HIGHEST EXPONENT.

First, we figure out which member of the denominator has the highest exponent.

Here is this n2,

but it is under a cube root.

Then, we have this n3,

but it doesn't have a chance, as it is under a 5th radical.

Finally, there is this n,

well, this seems to be the winner.

The member in the denominator that has the highest exponent is n, so we divide by that.

But if we bring it under the radical sign, it goes through some magical changes.

So, under the various radical signs, we divide by different powers of n.

Here come some very funny sequences.

Here is the first one.

Here come some very funny sequences.

Here is the first one.

Well, this is not the most difficult limit of our lives.

Here we have another one.

And a third one.

Well, there is nothing funny so far.

The real thrill comes now.

Here is this limit.

Based on the previous examples, this is supposedly 1, again.

The problem is: it is not.

It is because the exponent is not a specific number, but the index of the sequence, so it changes all the time.

If the exponent is a specific number, then all we did so far is OK.

But if the exponent contains n, then we are dealing with an entirely different animal.

In this case the limit is not 1, but a rather unpleasant number.

To avoid looking at this unpleasant number so often, they called it "e".

If it is not 1 here, but some other number,

then the limit will change a little bit.

And there is one more thing.

At least when

Well, let's see a few examples for this.

Here is this limit, for instance:

which is based on the formula

But if this part here changes,

and so does the exponent,

then, we get the same thing again.

Here is another one:

So, limits where these are the same can be computed very easily.

The question is what happens if they are not the same.

Then, we have to do something to make them the same. We either turn this into 2n,

or this into n.

Let's turn this into n.

Let's simplify the fraction by 2. And now it is good.

Here we will transform the exponent.

There is such a thing as:

There are some cases that are a bit more exciting:

But these are still not as exciting as those that come next.

Limits where  appears here

and in the exponent, too,

can always be computed using these formulas.

The trick is to twist and turn things until they resemble these formulas.

Well, to achieve our goal, let's divide the numerator and the denominator by .

If, by any chance, there is  in both the numerator and the denominator, then we divide by :

And we can fix uglier cases, too

If the exponent is a specific number, then:

But if, unfortunately, here we have  

and there  

then we only divide by , and then factor out

Sometimes there is n2 there.

Furthermore, it could be n3.

There is 2n3 at the top and also at the bottom, so let's divide by 2n3.

Here come some very funny sequences.

Here is the first one.

Well, this is not the most difficult limit of our lives.

Here we have another one.

And a third one.

Well, there is nothing funny so far.

The real thrill comes now.

Here is this limit.

Based on the previous examples, this is supposedly 1, again.

The problem is: it is not.

It is because the exponent is not a specific number, but the index of the sequence, so it changes all the time.

If the exponent is a specific number, then all we did so far is OK.

But if the exponent contains n, then we are dealing with an entirely different animal.

In this case the limit is not 1, but a rather unpleasant number.

To avoid looking at this unpleasant number so often, they called it "e".

If it is not 1 here, but some other number,

then the limit will change a little bit.

And there is one more thing.

At least when

Well, let's see a few examples for this.

Here is this limit, for instance:

which is based on the formula

But if this part here changes,

and so does the exponent,

then, we get the same thing again.

Here is another one:

So, limits where these are the same can be computed very easily.

The question is what happens if they are not the same.

Then, we have to do something to make them the same. We either turn this into 2n,

or this into n.

Let's turn this into n.

Let's simplify the fraction by 2. And now it is good.

Here we will transform the exponent.

There is such a thing as:

There are some cases that are a bit more exciting:

But these are still not as exciting as those that come next.

Limits where  appears here

and in the exponent, too,

can always be computed using these formulas.

The trick is to twist and turn things until they resemble these formulas.

Well, to achieve our goal, let's divide the numerator and the denominator by .

If, by any chance, there is  in both the numerator and the denominator, then we divide by :

And we can fix uglier cases, too

If the exponent is a specific number, then:

But if, unfortunately, here we have  

and there  

then we only divide by , and then factor out

Sometimes there is n2 there.

Furthermore, it could be n3.

There is 2n3 at the top and also at the bottom, so let's divide by 2n3.

A sequence is called convergent if there is a real number that is the limit of the sequence.

Convergent

sequences

Divergent

sequences

It has

a limit

It has

no limit

If there is no such number, then the sequence is divergent.

But there are degrees of divergence.

A sequence is divergent if it tends to infinity,

but it is also divergent if it doesn’t tend to anywhere at all.

Sequences that tend to nowhere are always oscillating sequences.

The simplest example of an oscillating sequence is the  sequence.

But before we start to think that all oscillating sequences are divergent, 

well, here comes another one.

So, just because a sequence bounces around, it isn’t necessarily divergent.

The size of those jumps is also important.

Here are three examples of the possible behaviors:

if n is even

if n is odd

And now let’s see some funny cases.

Well, here in the formula the 1 is up front, so let’s swap it.

It doesn’t pose a problem for addition.

For subtraction...

it is also OK, if we don’t mess it up.

And Voila! We got another oscillating sequence.

Now,  appears both in the numerator and the denominator,

so this is not an oscillating sequence.

And now let’s see some sequences with radicals.

Here comes another one.

Again, we identify which member of the denominator has the highest exponent.

And now let’s see something really interesting.

Well, there is nothing exciting yet.

The thrill comes when we replace the + sign...

with the  sign.

 is also , but

In such cases we need some magic.

From here, it is the usual drill.

And then we have this:

And one more:

If we have addition here, then we are done.

But if it is subtraction, then we need some hocus-pocus again.

A sequence is called convergent if there is a real number that is the limit of the sequence.

Convergent

sequences

Divergent

sequences

It has

a limit

It has

no limit

If there is no such number, then the sequence is divergent.

But there are degrees of divergence.

A sequence is divergent if it tends to infinity,

but it is also divergent if it doesn’t tend to anywhere at all.

Sequences that tend to nowhere are always oscillating sequences.

The simplest example of an oscillating sequence is the  sequence.

But before we start to think that all oscillating sequences are divergent, 

well, here comes another one.

So, just because a sequence bounces around, it isn’t necessarily divergent.

The size of those jumps is also important.

Here are three examples of the possible behaviors:

if n is even

if n is odd

And now let’s see some funny cases.

Well, here in the formula the 1 is up front, so let’s swap it.

It doesn’t pose a problem for addition.

For subtraction...

it is also OK, if we don’t mess it up.

And Voila! We got another oscillating sequence.

Now,  appears both in the numerator and the denominator,

so this is not an oscillating sequence.

And now let’s see some sequences with radicals.

Here comes another one.

Again, we identify which member of the denominator has the highest exponent.

And now let’s see something really interesting.

Well, there is nothing exciting yet.

The thrill comes when we replace the + sign...

with the  sign.

 is also , but

In such cases we need some magic.

From here, it is the usual drill.

And then we have this:

And one more:

If we have addition here, then we are done.

But if it is subtraction, then we need some hocus-pocus again.

Sequence  is “stronger” than sequence , if .

We write it like this:

You can visualize it as the stronger sequence tends to infinity faster.

For example, for large n values, n2 looks like this...

and n3 looks like this.

But when 2n, which is much stronger than both of them, enters the scene, ...

well, that significantly diminishes the self-esteem of the previous two sequences.

But 2n is not all-powerful, because 3n is much stronger.

It seems there is plenty of power-struggle in the world of sequences.

You’d better know everybody’s strength.

Now let's get down to business.

Here is a theorem that we will use for finding the limits of more complicated sequences.

The theorem says that if  

 and  and there is , such that for all  it holds that

 , then .

This theorem is called the Squeeze Theorem (or Sandwich Theorem), and one of its typical application area is finding

type limits.

Let's see this limit, for instance:

We have to realize that the strongest term here is , and

this means that for large n values the other terms are relatively tiny,

as if they weren’t there at all.

So, it shouldn’t be surprising that the limit

The Squeeze Theorem will help us to prove this in a precise manner.

Here is the plan:

First, we find a sequence that is less than the original sequence, 

and it has a limit of 5.

Next, we find another sequence that is greater than the original sequence, and also tends to 5.

Finally we conclude with content that the original sequence also tends to 5.

Well, this is a nice plan, but it isn’t that easy to carry out.

It is OK that we need a sequence that is less than the original sequence, 

and another one that is greater than the original sequence.

Anybody could come up with sequences like that. However, finding sequences that also tend to 5, well, that is a lot harder.

To do that, we need to learn the art of estimation.  

We have to be careful not to modify the strongest term during the estimation process.

When finding the lower estimate, we omit everything but the strongest term.

When finding the upper estimate, we replace all terms with the strongest term.

Let's see another limit.

The strongest term in the numerator is , and in the  denominator it is .

And now we are ready for the estimation.

To find the lower estimate, we decrease the numerator, and increase the denominator.

But we have to take care to not modify the strongest term neither in the numerator nor in the denominator.

We decrease the numerator by keeping only the strongest term.

We increase the denominator by replacing all terms with the strongest one.

When finding the upper estimate, we increase the numerator by replacing all terms with the strongest term.

We then decrease the denominator by keeping only the strongest term.

But there are more complicated cases, too:

The upper estimation is fairly easy, we just omit the negative term:

The lower estimation however, is quite cunning.

Here we need something that is less than the original.

So, it is not good to omit 5n, we should actually do the opposite: subtract more than that.

Yes, but what should we subtract? If we subtract - let’s say - 6n, then the lower estimate is zero.

And sadly, that is a problem...

We need to use a trick here.

Well, the lower estimate will be this. Here, C is some positive number, less than 1.

The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Let’s check whether it is less than the original:

Well, yes, if .

From here it is routine work.

Now, what’s next?

Now let’s continue with a few funny limits.

The strongest term is 6n, the others are irrelevant for the limit.

Since , the limit of the sequence is 6.

But this is only an intuition, an idea. For the precise proof, we need a theorem called the Squeeze Theorem.

And now we are ready for the precise solution.

Let’s start with the upper estimate, which happens to be very easy this time.

The lower estimate is more interesting.

We are going to do some hocus-pocus here. First, we replace each negative term with the strongest negative term.

And then we estimate it from below.

Here, C is some positive number, less than 1. The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Well, this holds if .

Both the lower and upper estimates tend to 6, so now the limit of the original sequence is officially confirmed as 6.

Well, this is wonderful news, so let’s check out one more.

The upper estimation is simple. We increase the numerator by omitting the negative terms,

and decrease the denominator by omitting this.

Since  is definitely not negative.

The lower estimate is more interesting.

We have to increase the denominator, and that is not a problem.

However, we have to decrease the numerator, and we will use the previous trick for that.

Well, this will hold sooner or later. To be exact, for n values greater than 8.

And now come a few very funny limits.

All of them will be like these:

On top of it, we will also have to use the Squeeze Theorem.

Let’s start with an easy one:

For the upper estimate, we decrease the denominator,

and for the lower estimate, we increase it:

Here comes a little trick.

Now these are the really funny cases.

We would be much happier without this n. As it turns out, that could actually be managed:

And Voila! From here it is exactly the same as the previous problem.

Sequence  is “stronger” than sequence , if .

We write it like this:

You can visualize it as the stronger sequence tends to infinity faster.

For example, for large n values, n2 looks like this...

and n3 looks like this.

But when 2n, which is much stronger than both of them, enters the scene, ...

well, that significantly diminishes the self-esteem of the previous two sequences.

But 2n is not all-powerful, because 3n is much stronger.

It seems there is plenty of power-struggle in the world of sequences.

You’d better know everybody’s strength.

Now let's get down to business.

Here is a theorem that we will use for finding the limits of more complicated sequences.

The theorem says that if  

 and  and there is , such that for all  it holds that

 , then .

This theorem is called the Squeeze Theorem (or Sandwich Theorem), and one of its typical application area is finding

type limits.

Let's see this limit, for instance:

We have to realize that the strongest term here is , and

this means that for large n values the other terms are relatively tiny,

as if they weren’t there at all.

So, it shouldn’t be surprising that the limit

The Squeeze Theorem will help us to prove this in a precise manner.

Here is the plan:

First, we find a sequence that is less than the original sequence, 

and it has a limit of 5.

Next, we find another sequence that is greater than the original sequence, and also tends to 5.

Finally we conclude with content that the original sequence also tends to 5.

Well, this is a nice plan, but it isn’t that easy to carry out.

It is OK that we need a sequence that is less than the original sequence, 

and another one that is greater than the original sequence.

Anybody could come up with sequences like that. However, finding sequences that also tend to 5, well, that is a lot harder.

To do that, we need to learn the art of estimation.  

We have to be careful not to modify the strongest term during the estimation process.

When finding the lower estimate, we omit everything but the strongest term.

When finding the upper estimate, we replace all terms with the strongest term.

Let's see another limit.

The strongest term in the numerator is , and in the  denominator it is .

And now we are ready for the estimation.

To find the lower estimate, we decrease the numerator, and increase the denominator.

But we have to take care to not modify the strongest term neither in the numerator nor in the denominator.

We decrease the numerator by keeping only the strongest term.

We increase the denominator by replacing all terms with the strongest one.

When finding the upper estimate, we increase the numerator by replacing all terms with the strongest term.

We then decrease the denominator by keeping only the strongest term.

But there are more complicated cases, too:

The upper estimation is fairly easy, we just omit the negative term:

The lower estimation however, is quite cunning.

Here we need something that is less than the original.

So, it is not good to omit 5n, we should actually do the opposite: subtract more than that.

Yes, but what should we subtract? If we subtract - let’s say - 6n, then the lower estimate is zero.

And sadly, that is a problem...

We need to use a trick here.

Well, the lower estimate will be this. Here, C is some positive number, less than 1.

The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Let’s check whether it is less than the original:

Well, yes, if .

From here it is routine work.

Now, what’s next?

Now let’s continue with a few funny limits.

The strongest term is 6n, the others are irrelevant for the limit.

Since , the limit of the sequence is 6.

But this is only an intuition, an idea. For the precise proof, we need a theorem called the Squeeze Theorem.

And now we are ready for the precise solution.

Let’s start with the upper estimate, which happens to be very easy this time.

The lower estimate is more interesting.

We are going to do some hocus-pocus here. First, we replace each negative term with the strongest negative term.

And then we estimate it from below.

Here, C is some positive number, less than 1. The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Well, this holds if .

Both the lower and upper estimates tend to 6, so now the limit of the original sequence is officially confirmed as 6.

Well, this is wonderful news, so let’s check out one more.

The upper estimation is simple. We increase the numerator by omitting the negative terms,

and decrease the denominator by omitting this.

Since  is definitely not negative.

The lower estimate is more interesting.

We have to increase the denominator, and that is not a problem.

However, we have to decrease the numerator, and we will use the previous trick for that.

Well, this will hold sooner or later. To be exact, for n values greater than 8.

And now come a few very funny limits.

All of them will be like these:

On top of it, we will also have to use the Squeeze Theorem.

Let’s start with an easy one:

For the upper estimate, we decrease the denominator,

and for the lower estimate, we increase it:

Here comes a little trick.

Now these are the really funny cases.

We would be much happier without this n. As it turns out, that could actually be managed:

And Voila! From here it is exactly the same as the previous problem.

Sequence  is “stronger” than sequence , if .

We write it like this:

You can visualize it as the stronger sequence tends to infinity faster.

For example, for large n values, n2 looks like this...

and n3 looks like this.

But when 2n, which is much stronger than both of them, enters the scene, ...

well, that significantly diminishes the self-esteem of the previous two sequences.

But 2n is not all-powerful, because 3n is much stronger.

It seems there is plenty of power-struggle in the world of sequences.

You’d better know everybody’s strength.

Now let's get down to business.

Here is a theorem that we will use for finding the limits of more complicated sequences.

The theorem says that if  

 and  and there is , such that for all  it holds that

 , then .

This theorem is called the Squeeze Theorem (or Sandwich Theorem), and one of its typical application area is finding

type limits.

Let's see this limit, for instance:

We have to realize that the strongest term here is , and

this means that for large n values the other terms are relatively tiny,

as if they weren’t there at all.

So, it shouldn’t be surprising that the limit

The Squeeze Theorem will help us to prove this in a precise manner.

Here is the plan:

First, we find a sequence that is less than the original sequence, 

and it has a limit of 5.

Next, we find another sequence that is greater than the original sequence, and also tends to 5.

Finally we conclude with content that the original sequence also tends to 5.

Well, this is a nice plan, but it isn’t that easy to carry out.

It is OK that we need a sequence that is less than the original sequence, 

and another one that is greater than the original sequence.

Anybody could come up with sequences like that. However, finding sequences that also tend to 5, well, that is a lot harder.

To do that, we need to learn the art of estimation.  

We have to be careful not to modify the strongest term during the estimation process.

When finding the lower estimate, we omit everything but the strongest term.

When finding the upper estimate, we replace all terms with the strongest term.

Let's see another limit.

The strongest term in the numerator is , and in the  denominator it is .

And now we are ready for the estimation.

To find the lower estimate, we decrease the numerator, and increase the denominator.

But we have to take care to not modify the strongest term neither in the numerator nor in the denominator.

We decrease the numerator by keeping only the strongest term.

We increase the denominator by replacing all terms with the strongest one.

When finding the upper estimate, we increase the numerator by replacing all terms with the strongest term.

We then decrease the denominator by keeping only the strongest term.

But there are more complicated cases, too:

The upper estimation is fairly easy, we just omit the negative term:

The lower estimation however, is quite cunning.

Here we need something that is less than the original.

So, it is not good to omit 5n, we should actually do the opposite: subtract more than that.

Yes, but what should we subtract? If we subtract - let’s say - 6n, then the lower estimate is zero.

And sadly, that is a problem...

We need to use a trick here.

Well, the lower estimate will be this. Here, C is some positive number, less than 1.

The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Let’s check whether it is less than the original:

Well, yes, if .

From here it is routine work.

Now, what’s next?

Now let’s continue with a few funny limits.

The strongest term is 6n, the others are irrelevant for the limit.

Since , the limit of the sequence is 6.

But this is only an intuition, an idea. For the precise proof, we need a theorem called the Squeeze Theorem.

And now we are ready for the precise solution.

Let’s start with the upper estimate, which happens to be very easy this time.

The lower estimate is more interesting.

We are going to do some hocus-pocus here. First, we replace each negative term with the strongest negative term.

And then we estimate it from below.

Here, C is some positive number, less than 1. The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Well, this holds if .

Both the lower and upper estimates tend to 6, so now the limit of the original sequence is officially confirmed as 6.

Well, this is wonderful news, so let’s check out one more.

The upper estimation is simple. We increase the numerator by omitting the negative terms,

and decrease the denominator by omitting this.

Since  is definitely not negative.

The lower estimate is more interesting.

We have to increase the denominator, and that is not a problem.

However, we have to decrease the numerator, and we will use the previous trick for that.

Well, this will hold sooner or later. To be exact, for n values greater than 8.

And now come a few very funny limits.

All of them will be like these:

On top of it, we will also have to use the Squeeze Theorem.

Let’s start with an easy one:

For the upper estimate, we decrease the denominator,

and for the lower estimate, we increase it:

Here comes a little trick.

Now these are the really funny cases.

We would be much happier without this n. As it turns out, that could actually be managed:

And Voila! From here it is exactly the same as the previous problem.