Uploaded by TheIntegralCALC on 31.10.2011

Transcript:

Hi, everyone. Welcome back to integralcalc.com. Today we’re going to be talking about how

to use the alternating series test to determine whether or not an alternating series converges.

And in this particular case, we have the sum of the series negative 1 to the k plus 1 times

the square root of k plus 1 minus the square root of k. And we’ve been asked to determine

whether or not this series from k equals 1 to infinity converges or not.

So because we have an alternating series, we’re going to use the alternating series

test for convergence. And I’ve taken here a screenshot of the part of the convergence

test chart that I have up on my website that talks about the alternating series test. So

what we see here is that in order to conclude that an alternating series converges, we have

to show two things. One, we need to show that the series a sub n is always greater than

the series a sub n plus 1 which is also always greater than zero. And we need to show that

the limit as n approaches infinity of the series a sub n is equal to zero. So those

are going to be the two things that we need to do and I’ll call them 1 and 2.

So we’ll start with the first piece, piece number 1 where we need to show that the series

a sub n is greater than the series a sub n plus 1 which is greater than zero. notice

here that in our formula for the alternating series, we have here a sub n. We’ve got

a sub n and then we also have this piece here which is the alternating series part of the

series. What this tells us is that our series is divided into two pieces. The first part

which gives us the alternating series part is going to be the negative one raised to

some exponent whether or not it’s k or k minus 1 or k plus 1. That’s the alternating

series part. Then the second part here, everything after that is going to be our series, a sub

n. So this second part right here is going to be a sub n. So we’ll use that as a sub

n. We need to show that that series a sub n is larger than the series a sub n plus 1.

So I’m going to show you two ways to do that. The first is kind of just to give you

an idea on your head of whether or not it is greater. The second is prove that it’s

greater. The first one I’m going to show you is that

the series a sub n is equal to the square root of k plus 1 minus the square root of

k. So we have that as the a sub n series. If we start plugging in for k and we start

with one because they tell us we’re starting with one here, we’ll get the square root

of 2 minus the square root of 1. Then if we plug in 2, we’ll get the square root of

3 minus the square root of 2. Then when we plug in 3, we’ll get the square root of

4 minus the square root of 3 and as you can see here the square root of 5 minus the square

root of 4. So those are going to be the terms we’re going to get when we plug in 1, 2,

3 and 4 on into infinity. So we’ll plug those values in for k. Then if we look at

a sub n plus 1, the first thing that we can try is just to replace k with k plus 1. So

we’ll say k plus 1 is going to go in for k so then we’ll have another plus 1 and

then we’re going to plug in k plus 1 here for the second k. So when we do that, obviously

we get the square root of k plus 2 minus the square root of k plus 1. So when we do that,

if we plug in again 1, 2, 3, and 4 starting with 1, we’ll get the square root of 3 minus

the square root of 2 plus the square root of 4 minus the square root of 3 when we plug

in 2 plus the square root of 5 minus the square root of 4, which is what we get when we plug

in 3 and then obviously the square root of 6 minus the square root of 5 when we plug

in 4. What we can see is that when we compare this first term here, if we compare the square

root of 2 minus the square root of 1, the square root of 3 minus the square root of

2, and then we compare every term in the series, when we plug in two, we got this term and

we got this term. When we compare each one of those in the series, what we find is that

a sub n, these terms appear are always greater than the a sub n plus 1 terms. So we conclude

that a sub n is greater than a sub n plus 1, what we get when we plug in k plus 1 for

k. Which is what we want up here. We want a sub n to be greater than a sub n plus 1.

So that’s how we can get an intuitive gut feel that this is going to work. But what

we want to do is prove that a sub n is greater than a sub n plus 1. And the way that we can

prove it is by taking the derivative of a sub n. If we take the derivative, it’s going

to tell us whether or not our series is increasing or decreasing. And if we can show that the

derivative is negative, the series is always decreasing, then of course a sub n plus 1,

a term further along the series will always be less than a sub n which is what we want

to show. So I’ll go ahead and show you that now. If you take the derivative of a sub n,

we’ll say d/dx of a sub n, the derivative of a sub n is going to be, and let’s go

ahead and transform this here into k plus 1 to the 1/2 minus k to the 1/2 which is

the same thing as keeping the square roots in there. The square root is the same thing

as taking what’s under the square root and raising it to the 1/2 power. So when we take

the derivative, we’ll multiply that 1/2 out in front, k plus 1 to the negative ½

and then we’ll get minus 1/2 times k to the negative 1/2. So when we simplify this,

we’ll end up with 1 over 2 times the square root of k plus 1 which we get because we move

this whole term here, k plus 1 to the negative 1/2. We move that to the denominator to make

the 1/2 exponent positive. When we move it from the numerator to the denominator, the

exponent becomes positive. And then of course, were back at positive 1/2 which is the same

thing as the square root. Similarly, over here, we’ve got 1 over 2 times the square

root of k. Now if we want to find a common denominator

here, we would just go ahead and multiply this first term times the square root of k

over the square root of k. And we’d multiply the second here by the square root of k plus

1, over the square root of k plus 1. When we simplify this, we’ll end up with the

square root of k minus the square root of k plus 1 all over 2 times the square root

of k times the square root of k plus 1. So now, to simplify further, what we can do is

multiply by the conjugate. Remember from limits, we can multiply by the conjugate of the numerator

here. It’s the same two terms but instead of the negative sign in between them, we put

a positive sign. This term right here is the conjugate of the term from our original numerator.

So k plus k plus 1. We multiply the numerator and denominator by the conjugate so that it’s

the same thing as multiplying by one. And when we now multiply these out, we’ll get

k and then we have plus the square root of k times the square root of k plus 1 minus

the square root of k times the square root of k plus 1. So that cancels. Then when we

multiply these two terms together, we’re just left with minus k plus 1. So we get that

together and then in the denominator here, let’s combine these two. you can combine

those by putting everything underneath the square root sign when two square roots are

multiplied together so we’ll just do k squared plus k. And then we have times the square

root of k plus the square root of k plus 1. Alright. So now you can see that we’ve got

a k and a minus k. So these two are going to cancel and all we’re left with here and

let me just go ahead and cancel this whole thing. All we’re left with in the numerator

is a negative 1. So this is our derivative, d/dx of a sub n. This is our derivative and

what you can see is that the numerator is always going to be negative. The denominator

is always going to be positive because our series starts with k equals 1. If you plug

in 1 or any number larger than 1 in for k into our denominator here, obviously, you’re

going to get positive 2 there. You’re going to get positive underneath the square root

sign here because you’ll get positive number squared plus another positive number is always

going to give you a positive number underneath that square root sign. So everything in the

denominator is positive. So since you’ve got a negative divided by a positive, that’s

always going to be a negative which means that the derivative, d/dx of a sub n is always

going to be negative and so it’s always going to be less than zero which means that

the function is always decreasing. So if the function is always decreasing, our series

basically goes a sub n plus a sub n plus 1 plus a sub n plus 2, etc.

Since the series is always decreasing, a sub n plus 1 will always give us an a sub n, a

sub n plus 2 will always be less than a sub n plus 1, and the series will be constantly

decreasing. So we’ve now proven here that a sub n is greater always than a sub n plus

1. We also know that a sub n plus 1 is always going to be greater than zero and w can see

that over here in our series that we’ve written out. No matter how large you go, four

hundred terms out or whatever, you’re going to have the square root of 400 minus the square

root of 399, that’s always still going to give you a positive number. It will always

be positive in there. It will always be greater than zero. So we have now officially checked

off these two requirements. We’ve proven there that a sub n will always be greater

than a sub n plus 1 which will always be greater than zero. So that’s how you can prove using

the derivative that you’ve met that first requirement.

So now all we need to do is determine whether or not the limit of a sub n as n goes to infinity

is equal to zero, number 2. If we can prove that, then we know that this series converges.

So let’s go ahead and look at that. So to show that the limit is equal to zero,

what we’ll go ahead and do is take the series a sub n, so remember that the series is the

square root of k plus 1 minus the square root of k. And what we want to do is find the limit

of the series as n approaches infinity. And we want it to be equal to zero. We can see,

roughly, just by looking at it that it’s probably going to be equal to zero. If we

plug in infinity for k right here, we’ll get infinity plus 1 which one compared to

infinity is nothing. So it’s basically infinity. And then we get infinity here, square root

of infinity minus infinity should be zero. But the way that we can prove this is by multiplying

again by the conjugate and that is allowing us to see very clearly that the limit is going

to be zero as k approaches infinity. So we’re multiplying another conjugate which is the

square root of k plus 1 plus square root of k. So when we multiply by the conjugate, we’ll

see that we’ll get k plus 1. Then we’ll have plus square root of k times square root

of k plus 1 minus square root of k times square root of k plus 1, which will cancel each other

out and then we’ll just be left with a minus k. So that will be our numerator and then

in our denominator, we’ll end up with square root of k plus 1 plus square root of k. And

we want to be finding of course the limit as k goes to infinity.

So we can see obviously that k and k here are going to cancel so we’ll be left with

1 over the square root of k plus 1 plus the square root of k and this should make things

very clear for us because obviously our denominator now when we plug in infinity here for k, our

whole denominator is going to be very, very, very large, infinitely large. And we know

from doing lots of these problems that whenever the denominator gets very large, you have

1 over some very large number or anything really over some very large number, the answer

here is going to be zero because 1 divided by an extremely large number is always going

to tend toward zero. So the limit as k approaches infinity of a sub k, which is our series a

sub k or a sub n, is going to be equal to zero and so now, we can say that we have proven

the second portion of these requirement. Both of these requirements we’ve checked off

and therefore, we’ve used the alternating series test to prove this particular series

converges. So that’s it. I hope this video helped you

guys and I will see you in the next one. Bye!

to use the alternating series test to determine whether or not an alternating series converges.

And in this particular case, we have the sum of the series negative 1 to the k plus 1 times

the square root of k plus 1 minus the square root of k. And we’ve been asked to determine

whether or not this series from k equals 1 to infinity converges or not.

So because we have an alternating series, we’re going to use the alternating series

test for convergence. And I’ve taken here a screenshot of the part of the convergence

test chart that I have up on my website that talks about the alternating series test. So

what we see here is that in order to conclude that an alternating series converges, we have

to show two things. One, we need to show that the series a sub n is always greater than

the series a sub n plus 1 which is also always greater than zero. And we need to show that

the limit as n approaches infinity of the series a sub n is equal to zero. So those

are going to be the two things that we need to do and I’ll call them 1 and 2.

So we’ll start with the first piece, piece number 1 where we need to show that the series

a sub n is greater than the series a sub n plus 1 which is greater than zero. notice

here that in our formula for the alternating series, we have here a sub n. We’ve got

a sub n and then we also have this piece here which is the alternating series part of the

series. What this tells us is that our series is divided into two pieces. The first part

which gives us the alternating series part is going to be the negative one raised to

some exponent whether or not it’s k or k minus 1 or k plus 1. That’s the alternating

series part. Then the second part here, everything after that is going to be our series, a sub

n. So this second part right here is going to be a sub n. So we’ll use that as a sub

n. We need to show that that series a sub n is larger than the series a sub n plus 1.

So I’m going to show you two ways to do that. The first is kind of just to give you

an idea on your head of whether or not it is greater. The second is prove that it’s

greater. The first one I’m going to show you is that

the series a sub n is equal to the square root of k plus 1 minus the square root of

k. So we have that as the a sub n series. If we start plugging in for k and we start

with one because they tell us we’re starting with one here, we’ll get the square root

of 2 minus the square root of 1. Then if we plug in 2, we’ll get the square root of

3 minus the square root of 2. Then when we plug in 3, we’ll get the square root of

4 minus the square root of 3 and as you can see here the square root of 5 minus the square

root of 4. So those are going to be the terms we’re going to get when we plug in 1, 2,

3 and 4 on into infinity. So we’ll plug those values in for k. Then if we look at

a sub n plus 1, the first thing that we can try is just to replace k with k plus 1. So

we’ll say k plus 1 is going to go in for k so then we’ll have another plus 1 and

then we’re going to plug in k plus 1 here for the second k. So when we do that, obviously

we get the square root of k plus 2 minus the square root of k plus 1. So when we do that,

if we plug in again 1, 2, 3, and 4 starting with 1, we’ll get the square root of 3 minus

the square root of 2 plus the square root of 4 minus the square root of 3 when we plug

in 2 plus the square root of 5 minus the square root of 4, which is what we get when we plug

in 3 and then obviously the square root of 6 minus the square root of 5 when we plug

in 4. What we can see is that when we compare this first term here, if we compare the square

root of 2 minus the square root of 1, the square root of 3 minus the square root of

2, and then we compare every term in the series, when we plug in two, we got this term and

we got this term. When we compare each one of those in the series, what we find is that

a sub n, these terms appear are always greater than the a sub n plus 1 terms. So we conclude

that a sub n is greater than a sub n plus 1, what we get when we plug in k plus 1 for

k. Which is what we want up here. We want a sub n to be greater than a sub n plus 1.

So that’s how we can get an intuitive gut feel that this is going to work. But what

we want to do is prove that a sub n is greater than a sub n plus 1. And the way that we can

prove it is by taking the derivative of a sub n. If we take the derivative, it’s going

to tell us whether or not our series is increasing or decreasing. And if we can show that the

derivative is negative, the series is always decreasing, then of course a sub n plus 1,

a term further along the series will always be less than a sub n which is what we want

to show. So I’ll go ahead and show you that now. If you take the derivative of a sub n,

we’ll say d/dx of a sub n, the derivative of a sub n is going to be, and let’s go

ahead and transform this here into k plus 1 to the 1/2 minus k to the 1/2 which is

the same thing as keeping the square roots in there. The square root is the same thing

as taking what’s under the square root and raising it to the 1/2 power. So when we take

the derivative, we’ll multiply that 1/2 out in front, k plus 1 to the negative ½

and then we’ll get minus 1/2 times k to the negative 1/2. So when we simplify this,

we’ll end up with 1 over 2 times the square root of k plus 1 which we get because we move

this whole term here, k plus 1 to the negative 1/2. We move that to the denominator to make

the 1/2 exponent positive. When we move it from the numerator to the denominator, the

exponent becomes positive. And then of course, were back at positive 1/2 which is the same

thing as the square root. Similarly, over here, we’ve got 1 over 2 times the square

root of k. Now if we want to find a common denominator

here, we would just go ahead and multiply this first term times the square root of k

over the square root of k. And we’d multiply the second here by the square root of k plus

1, over the square root of k plus 1. When we simplify this, we’ll end up with the

square root of k minus the square root of k plus 1 all over 2 times the square root

of k times the square root of k plus 1. So now, to simplify further, what we can do is

multiply by the conjugate. Remember from limits, we can multiply by the conjugate of the numerator

here. It’s the same two terms but instead of the negative sign in between them, we put

a positive sign. This term right here is the conjugate of the term from our original numerator.

So k plus k plus 1. We multiply the numerator and denominator by the conjugate so that it’s

the same thing as multiplying by one. And when we now multiply these out, we’ll get

k and then we have plus the square root of k times the square root of k plus 1 minus

the square root of k times the square root of k plus 1. So that cancels. Then when we

multiply these two terms together, we’re just left with minus k plus 1. So we get that

together and then in the denominator here, let’s combine these two. you can combine

those by putting everything underneath the square root sign when two square roots are

multiplied together so we’ll just do k squared plus k. And then we have times the square

root of k plus the square root of k plus 1. Alright. So now you can see that we’ve got

a k and a minus k. So these two are going to cancel and all we’re left with here and

let me just go ahead and cancel this whole thing. All we’re left with in the numerator

is a negative 1. So this is our derivative, d/dx of a sub n. This is our derivative and

what you can see is that the numerator is always going to be negative. The denominator

is always going to be positive because our series starts with k equals 1. If you plug

in 1 or any number larger than 1 in for k into our denominator here, obviously, you’re

going to get positive 2 there. You’re going to get positive underneath the square root

sign here because you’ll get positive number squared plus another positive number is always

going to give you a positive number underneath that square root sign. So everything in the

denominator is positive. So since you’ve got a negative divided by a positive, that’s

always going to be a negative which means that the derivative, d/dx of a sub n is always

going to be negative and so it’s always going to be less than zero which means that

the function is always decreasing. So if the function is always decreasing, our series

basically goes a sub n plus a sub n plus 1 plus a sub n plus 2, etc.

Since the series is always decreasing, a sub n plus 1 will always give us an a sub n, a

sub n plus 2 will always be less than a sub n plus 1, and the series will be constantly

decreasing. So we’ve now proven here that a sub n is greater always than a sub n plus

1. We also know that a sub n plus 1 is always going to be greater than zero and w can see

that over here in our series that we’ve written out. No matter how large you go, four

hundred terms out or whatever, you’re going to have the square root of 400 minus the square

root of 399, that’s always still going to give you a positive number. It will always

be positive in there. It will always be greater than zero. So we have now officially checked

off these two requirements. We’ve proven there that a sub n will always be greater

than a sub n plus 1 which will always be greater than zero. So that’s how you can prove using

the derivative that you’ve met that first requirement.

So now all we need to do is determine whether or not the limit of a sub n as n goes to infinity

is equal to zero, number 2. If we can prove that, then we know that this series converges.

So let’s go ahead and look at that. So to show that the limit is equal to zero,

what we’ll go ahead and do is take the series a sub n, so remember that the series is the

square root of k plus 1 minus the square root of k. And what we want to do is find the limit

of the series as n approaches infinity. And we want it to be equal to zero. We can see,

roughly, just by looking at it that it’s probably going to be equal to zero. If we

plug in infinity for k right here, we’ll get infinity plus 1 which one compared to

infinity is nothing. So it’s basically infinity. And then we get infinity here, square root

of infinity minus infinity should be zero. But the way that we can prove this is by multiplying

again by the conjugate and that is allowing us to see very clearly that the limit is going

to be zero as k approaches infinity. So we’re multiplying another conjugate which is the

square root of k plus 1 plus square root of k. So when we multiply by the conjugate, we’ll

see that we’ll get k plus 1. Then we’ll have plus square root of k times square root

of k plus 1 minus square root of k times square root of k plus 1, which will cancel each other

out and then we’ll just be left with a minus k. So that will be our numerator and then

in our denominator, we’ll end up with square root of k plus 1 plus square root of k. And

we want to be finding of course the limit as k goes to infinity.

So we can see obviously that k and k here are going to cancel so we’ll be left with

1 over the square root of k plus 1 plus the square root of k and this should make things

very clear for us because obviously our denominator now when we plug in infinity here for k, our

whole denominator is going to be very, very, very large, infinitely large. And we know

from doing lots of these problems that whenever the denominator gets very large, you have

1 over some very large number or anything really over some very large number, the answer

here is going to be zero because 1 divided by an extremely large number is always going

to tend toward zero. So the limit as k approaches infinity of a sub k, which is our series a

sub k or a sub n, is going to be equal to zero and so now, we can say that we have proven

the second portion of these requirement. Both of these requirements we’ve checked off

and therefore, we’ve used the alternating series test to prove this particular series

converges. So that’s it. I hope this video helped you

guys and I will see you in the next one. Bye!