Hi, everyone.
Welcome back to integralcalc.com.
Today, we’re going to be talking about finding the sum of a telescoping series.
And in this particular problem, we’ve been asked to find the sum from n equals 1 to infinity
of 1 divided by 4n squared minus 1.
And with any telescoping series and finding the sum of a telescoping series, what you’re
going to want to do is perform a partial fractions decomposition on the fraction you’ve been
given for your sum.
So that means that you’re going to want to factor our denominator as much as we can.
So if we factor our denominator, 4n squared minus 1, we’ll find that it’s equal to
2n plus 1 times 2n minus 1.
And remember, to perform a partial fractions decomposition, we’re going to set this fraction
equal to a divided by the first term on the denominator on the left so 2n plus 1 and then
we’re going to put b over the other term, so we’ll add and then we’ll get b over
2n minus 1.
And now, we’re going to want to multiply both sides by the denominator from the left-hand
side.
So we’re going to multiply this whole thing by 2n plus 1 times 2n minus 1.
And as you can tell, what that’s going to do is cancel the entire denominator over here
on the left so we’ll just be left with 1 equals.
For the term that involves a here, the 2n plus 1 is going to cancel with this 2n plus
1 and we’ll be left with 2n minus 1.
And then for the second fraction with b, the 2n minus 1 will cancel with 2n minus 1 and
we’ll be left with b times 2n plus 1.
Now, it’s just a matter of simplifying this equation and then developing a system of simultaneous
equations that we can solve for the coefficients a and b.
So we’re just going to multiply things out here and we’ll get 2an minus a plus 2bn
plus b.
And then we’ll go ahead and group our terms together so we’ll say 2an plus 2bn.
We’re grouping those together because they both have an n to the first power variable
involved with them.
And then we’re going to group together negative a and positive b because neither of those
terms have an n variable involved.
They are the constants in this equation.
So now, we’ll go ahead and factor out the n.
So we’ll get 2a plus 2b, we’re just factoring out the n so we pull that out there, plus
negative a plus b.
So now we can equate coefficients on the left and right-hand sides and hopefully you’ll
remember this from partial fractions but basically on the left-hand side, we have zero n plus
1.
There’s no term that involves the variable n on the left-hand side so we have zero n
plus 1.
1 is the constant, zero n is the first degree term.
So because the coefficient over here on the left on the n term is zero, that means the
coefficient on the n term on the right-hand side has to equal zero.
So we grab those two and then we say 2a plus 2b must be equal to zero.
And then similarly, because the constant on the left is equal to one, that means that
the constant on the right, negative a plus b, must also be equal to 1.
So we’ll say negative a plus b equals 1.
So now let’s go ahead and solve these system of simultaneous equations.
We’ll take this second equation and we’ll add a to both sides to find that b is equal
to 1 plus a.
Now, we can take this value one plus a and plug it in for b into our first equation.
So we’ll find 2a plus 2 and here’s when we make our substitution for b of 1 plus a.
And that’s going to be equal to zero.
So now, we can multiply out.
We’ll get 2a plus 2 plus 2a equals zero.
We’ll find that 4a equals negative 2.
And therefore, the a is equal to negative 1/2.
Now that we know that a equals negative 1/2, we will plug that in to this equation for
b and we’ll get b equals 1 minus 1/2.
Therefore, b is equal to positive 1/2.
So we can note these two solutions for the constants a and b.
So once we have those solutions now that we’re finished with our partial fractions decomposition,
we can go ahead and plug them back in to this part of our equation right here and this is
what we’ll use to determine the series.
So we’ll plug in negative 1/2 for a.
So we’ll get negative 1/2 over 2n plus 1.
And then b is 1/2 so we’ll get plus 1/2 divided by 2n minus 1.
And let’s go ahead and flip this around so that we’re not leading with a negative.
We’ll end up with 1/2 over 2n minus 1 minus 1/2 over 2n plus 1.
So now, this is our series.
We’ll call this s sub n.
And what we want to do is go ahead and start plugging in a value for n.
So we’ve already been told in our problem that we need to start with a value of 1 when
we plug in for n.
So that’s what we’re going to do.
We’re going to say that some other series is going to be equal to and we’re going
to start writing out terms.
Obviously, this is an infinite series so we can’t write out every term but we’re going
to write out the first few to just give us an idea of what the series is starting to
look like.
So, let’s just say that we’re plugging in 1 for n here in our first term and we’re
going to plug it in to this whole series.
So in this first one, we’ll get 1/2 divided by 2 times 1 minus 1 minus 1/2 over 2 times
1 plus 1.
So as you can see in the bottom here, we’re going to get 1 so that’s going to be 1/2
over 1 which is just 1/2.
And over here on our second fraction, we’ll end up with three in the denominator. 1/2
divided by 3 is going to be 1/6.
So that means that our first term is going to be 1/2 minus 1/6.
So you can call this n equals 1.
Now, if we go ahead and plug in 2 for n, we’ll get 1/2 divided by 2 times 2 minus 1 minus
1/2 over 2 times 2 plus 1.
For the first fraction, we’ll get 1/2 divided by 4 minus 1 is 3 so that’s actually going
to give us a 1/6 and then we’ll subtract.
In the denominator here, we’ve got 2 times 2 which is 4 p;lus 1 is 5.
1/2 divided by 5 is 1/10.
So that’s now our second term.
1/6 minus 1/10.
And this is going to be n equals 2.
Let’s just do one more to get an idea.
So if we plug in 3, we’ll get 1/2 over 2 times 3 minus 1 minus 1/2 divided by 2 times
3 plus 1.
In the denominator over here, we’ll get 2 times 3 which is 6 minus 1 is 5.
1/2 divided by 5 is 1/10.
And then on the denominator here, we get 2 times 3 which is 6 plus 1 which is 7.
1/2 divided by 7 is 1/14.
So we end up for n equals 3 with 1/10 minus 1/14.
And this is what we get for n equals 3.
So at this point, we can start talking about what our series is shaping up to look like.
Basically, at this point, what you need to recognize is that as we go out toward infinity,
we’re going to end up with this interesting pattern where notice here we have negative
1/6 and then we have a positive 1/6.
Those two term will cancel one another.
We have a negative 1/10 and a positive 1/10 so those two are going to cancel.
If we found the term for n equals 4, we would end up with positive 1/14 minus 1/18 and this
1/14 term would cancel with the next one and we’d be able to continue canceling each
of these terms out into infinity.
And what that tells us is that all we’re left with is the very first half of our first
term 1/2.
And therefore, that the sum of the series s sub n is equal to 1/2.
Or that the sum of this series is equal to 1/2.
And that’s really it.
So that’s our final answer.
This is the sum of the telescoping series.
So remember, it’s really just a matter of performing your partial fractions decomposition,
getting this s sub n equation set up with your a and b constants, your partial fractions
decomposition.
Then you start plugging in values for n, you see that the series breaks out this way and
you’re able to cancel all of the terms except the very first one and therefore you know
that the sum of your series is just the first term, which in this case is 1/2.
So that’s it.
I really hope this video helped you guys and I’ll see you in the next one.
Bye!