### Video Transcript

Remainder of an Alternating
Series

In this video, weβre going to
discuss approximating certain alternating series by their partial sum. We will see the types of
alternating series we can approximate this way, and we will also see how to
calculate the error of this approximation. We will also see how to use this to
find an approximation up to a certain level of error that we want. Answering questions like, how many
terms do we need to approximate this alternating series up to a certain level of
error? Letβs start by discussing the types
of alternating series weβre going to approximate.

Let π π be a positive and
decreasing sequence where the limit as π approaches β of π π is equal to
zero. We chose these properties for π π
so that it satisfies the alternating series test. This means the sum from π equals
one to β of negative one to the power of π plus one times π π converges. Letβs call this value π. We know that this value is equal to
the limit of its partial sums. And since this limit converges,
this means the limit of its partial sums also converges. So, as we take more and more terms
in our partial sum, we get closer and closer to π. This means we can approximate π by
using the πth partial sum. We just take more and more terms to
get a more and more accurate representation of π.

But just how accurate will this sum
be? Letβs start by rewriting our
infinite alternating series in terms of the πth partial sum. Our infinite alternating series
will be equal to the πth partial sum plus all of the remaining terms. So, we have that our infinite
series is equal to the πth partial sum plus the remainder after π terms. We will call this π
π. And we want to know how close our
estimate of the partial sum is to the actual value of π. If we have that π is equal to the
πth partial sum plus the remainder after π terms, then by subtracting the πth
partial sum from both sides of this equation, we have that π minus π π is equal
to π
π. In other words, the difference
between the actual value of our series and our approximation is just equal to π
π. So, π
π is actually telling us
how accurate our approximation is.

We can actually rewrite our initial
equation in terms of power series to reflect this. The power series for π minus the
power series for the πth partial sum is equal to our remainder terms. So, to estimate how accurate our
approximation is, we just need to estimate the value of π
π. To find an approximation of π
π,
letβs write this series out term by term. We can see that this is an
alternating series. We can see that the sign of each
term alternates between positive and negative. In fact, if we assume π is even,
then π plus two is even, so negative one raised to this power is just one. So our first term, π π plus one,
is positive. And we need to alternate the signs
of the rest of the terms.

At first, it might not seem like
this will help us approximate the value of this series. However, we chose our sequence to
be decreasing. That means each term in this series
is getting smaller and smaller in absolute value. In fact, we also chose our sequence
to be positive. So, what does this tell us? Letβs compare the values of π π
plus three and negative π π plus two. We know that π π plus three is
smaller than π π plus two because our sequence is decreasing. This is the same as saying that
negative π π plus two plus π π plus three is less than zero. What weβve shown is if we pair off
these two terms in our series, we get a negative term.

We can also see this is true for
the next two terms in our series. π π plus five is smaller than π
π plus four, and these are both positive. So, pairing off these two terms in
our series leaves us with another negative number. But remember, the first term in our
sequence is positive, so our series is a positive number and then we add negative
numbers. This means we can bound it above by
π π plus one, since adding negative numbers to a positive number will make it
smaller. So, we found a bound for our
estimate when we take an even number of terms in our partial sum. But what would have happened if we
had taken an odd number of terms in our partial sum?

We could write out our series term
by term again. However, this time we would start
with negative π π plus one. We could then try pairing off the
same terms we did before. However, this time we see that π
π plus two is bigger than π π plus three. So, π π plus two minus π π plus
three must be positive. In fact, this is true for all of
these pairs. Weβre subtracting a smaller
positive number from a bigger positive number. And we see that the leading term of
this series, negative π π plus one, is negative. In other words, we start with a
negative number, and then we add positive terms. This makes our number bigger, so we
can say that this is greater than or equal to negative π π plus one.

And we can now see we have very
similar bounds in the case where π is even and the case where π is odd. When π was even, weβve shown that
π
π is less than or equal to π π plus one. And when π is odd, weβve shown
that π
π is greater than or equal to negative π π plus one. To combine these into one bound, we
first want to show π
π is positive when π is even. To do this, letβs go back to our
series expansion when π was even, except this time weβre going to pair our terms
this way.

And we can see, evaluating the
terms inside of the parentheses always leaves a positive number. So when π is even, π
π is the
sum of positive numbers, so itβs positive. Since π π plus one is also
positive, we can take the absolute value of both sides of this inequality. And this gives us the absolute
value of π
π is less than or equal to the first neglected term π π plus one.

We can do the same when π is
odd. We need to show that π
π is
negative. When π is odd, we write our series
out term by term. However, this time we pair our
terms in this way. And if we evaluate the expressions
inside of our parentheses, we see we always get a negative value. So when π is odd, π
π is
negative. Itβs the sum of negative terms.

We again want to take the absolute
value of both sides of this inequality. However, we must be careful. Negative π π plus one is
negative, and π
π is negative. So, when we take the absolute
value, we need to flip the direction of the inequality. And we can simplify the absolute
value of negative π π plus one. Itβs just π π plus one. So, in both of these cases, we can
bound the error of our approximation by the absolute value of the first neglected
term. Letβs see how we can use this to
approximate an alternating series.

Find the maximum error bound when
approximating the series the sum from π equals one to β of negative one to the πth
power multiplied by the square root of three π plus seven divided by π squared
plus one by summing the first 20 terms. Round your answer to five decimal
places.

The question wants us to find the
maximum possible error we could get by approximating this series by summing the
first 20 terms. It wants us to round this value to
five decimal places. The question wants us to
approximate this series by summing the first 20 terms. Thatβs the same as taking the 20th
partial sum. And the maximum possible error
bound would be a bound on the absolute value of π minus our 20th partial sum, where
π would be the value that our series converges to.

To help us find this bound, we know
that if π π is a positive and decreasing sequence where the limit as π approaches
β of π π is equal to zero, then, by the alternating series test, the alternating
series the sum from π equals one to β of negative one to the πth power times π π
converges β weβll call this value π.

Then, we can bound the error
between the value of π and our πth partial sum by using the absolute value of π
minus the πth partial sum is less than or equal to π π plus one, the absolute
value of the first neglected term. By looking at the series given to
us in the question, weβll want to set π π equal to the square root of three π
plus seven divided by π squared plus one. If we can then show that π π is a
positive decreasing sequence whose limit as π approaches β is equal to zero, then,
by setting π equal to 20, weβll have the absolute value of π minus the 20th
partial sum is less than or equal to π 21.

Letβs start by showing that π π
is positive. We know that π is greater than or
equal to one. So, three π plus seven is
positive, and then π squared plus one is positive. And weβre taking the positive
square root. So, our sequence π π is positive
for all values of π since weβre just taking the positive square root of a positive
number.

To check that the sequence is
decreasing, weβll set π of π₯ equal to the square root of three π₯ plus seven
divided by π₯ squared plus one. We know that this sequence will be
decreasing if the slope of this function is negative. To help us differentiate this
function, weβll start by setting π’ equal to three π₯ plus seven divided by π₯
squared plus one and then using the chain rule. Since π is a function of π’ and π’
is a function of π₯, by the chain rule we have π prime of π₯ is equal to the
derivative of the square root of π’ with respect to π’, which we can evaluate by
using the power rule for differentiation. This gives us a half multiplied by
π’ to the power of negative a half. And then, we need to multiply this
by dπ’ by dπ₯.

Remember, we only need to calculate
whether the slope is negative or positive. Since weβre only interested in the
values of π₯ where π₯ is greater than or equal to one, we can see that π’ as
positive; itβs the quotient of two positive numbers. So, a half is positive, and π’ to
the power of negative a half is also positive. Itβs one divided by the positive
square root of a positive number. So, to decide whether the slope is
positive or negative, we only need to calculate dπ’ by dπ₯. To find dπ’ by dπ₯, weβll use the
quotient rule. Weβll set π£ equal to the
numerator, three π₯ plus seven, and π€ equal to the denominator, π₯ squared plus
one. This gives us π£ prime is three and
π€ prime is two π₯.

The quotient rule then tells us dπ’
by dπ₯ is equal to π£ prime π€ minus π£π€ prime all divided by π€ squared. This gives us three times π₯
squared plus one minus three π₯ plus seven times two π₯ all divided by π₯ squared
plus one all squared. We can evaluate the numerator to
get three π₯ squared plus three minus six π₯ squared plus 14π₯, which we can
simplify to give us negative three π₯ squared minus 14π₯ plus three all divided by
π₯ squared plus one squared.

And we see, for values of π₯
greater than or equal to one, π₯ squared plus one all squared is positive. However, negative three π₯ squared
minus 14π₯ plus three is negative. So, dπ’ by the dπ₯ is negative for
these values of π₯. This means that our slope, π prime
of π₯, is a positive number multiplied by a negative number, which means π prime of
π₯ is negative. And if our slope is negative for
these values of π₯, our sequence must be decreasing.

We now want to check the limit as
π approaches β of π π is equal to zero. We see that π π is the square
root of a fraction. Weβll divide both the numerator and
the denominator of this fraction by the highest power of π which appears in the
fraction. Thatβs π squared. Dividing through by π squared
gives us the limit as π approaches β of the square root of three over π plus seven
over π squared all divided by one plus one over π squared.

We can see as π is approaching β,
our numerator of three over π plus seven over π squared is approaching zero. And we can see the one over π
squared term in our denominator is also approaching zero. However, the term one remains
constant. So, this fraction is approaching
zero divided by one; itβs approaching zero. And since the limit of a power is
equal to the power of the limit, this means that this limit is approaching zero.

So, weβve shown as π approaches β,
π π approaches zero. This means the maximum error bound
when summing the first 20 terms of our series is just π 21. So, if we were to approximate this
series by summing the first 20 terms, our error would be at most π 21, which is
equal to the square root of three times 21 plus seven divided by 21 squared plus
one. And if we calculate this to five
decimal places, we get 0.39796.

So, weβve now seen how to use this
to find the maximum possible error. Letβs see how we can now use this
to estimate the value of a series up to a certain level of accuracy.

Calculate the partial sum π π for
the least π terms which guarantees that the sum of the first π terms of the
alternating series the sum from π equals one to β of negative one to the power of
π plus one divided by five to the πth power differs from the infinite sum by 10 to
the power of negative six at most. Give your answer to six decimal
places.

The question wants us to
approximate this alternating series by using a partial sum. It wants us to use an approximation
which uses the least number of terms which guarantees that our difference between
our estimate and the infinite sum is at most 10 to the power of negative six. Once we found this value of π, we
need to calculate the partial sum to six decimal places. We see that the series given to us
is a geometric series with initial term π equal to one-fifth and ratio of
successive terms, π, equal to negative one-fifth. So, in this case, we could actually
just calculate the value of the infinite sum. Then, all we would need to do is
add more and more terms to our partial sums until weβre within 10 to the power of
negative six of the infinite sum. And this would work.

However, we donβt know how many
terms we would need to add. This could take hundreds and
hundreds of terms for all we know. Instead, weβre going to try and
approximate how many terms we would need. To help us estimate the value of π
we need, we recall the following fact about alternating series. If π π is a positive and
decreasing sequence whose limit as π approaches β is equal to zero, then the
alternating series the sum from π equals one to β of negative one to the power of
π plus one times π π is convergent by the alternating series test β weβll call
this equal to π. Then, we know we can bound the
absolute value of π minus the πth partial sum by the first term we left out. Itβs less than or equal to π π
plus one. We will set π π equal to one
divided by five to the πth power.

We see if we set π equal to
one-fifth and π also equal to one-fifth, then we can see that our sequence π π is
also a geometric sequence. This actually then tells us all the
information we need to know. One-fifth to the πth power is
always positive. And the absolute value of one-fifth
is less than one, so itβs decreasing. Finally, we know the limit as π
approaches β of one divided by five to the πth power is equal to zero. So, we can use this to approximate
the difference between the πth partial sum and the actual value of our infinite
series π. The absolute value of π minus the
πth partial sum is less than or equal to one divided by five to the power of π
plus one.

Remember, we want this error to be
at most 10 to the power of negative six. So, if we chose a value of π such
that one divided by five to the power of π plus one was less than or equal to 10 to
the power of negative six, then, for this particular value of π, the absolute value
of π minus the πth partial sum is less than or equal to 10 to the power of
negative six. So, this value of π is
sufficient. However, itβs not necessarily the
lowest value of π. So, letβs find the sufficient value
of π. We want one divided by five to the
power of π plus one to be less than or equal to 10 to the power of negative
six.

Both of these terms are
positive. So, we take the reciprocal of both
sides of this equation and then flip the inequality. We then take the log base five of
both sides of this inequality. Then, we just subtract one from
both sides of the inequality. Evaluating this gives us π is
greater than or equal to 7.6, which is the same as saying π is greater than or
equal to eight. So, what we have done is we have
shown the absolute value of π minus the eighth partial sum is less than or equal to
one divided by five to the power of eight plus one, which itself is less than or
equal to 10 to the power of negative six. In other words, when π is equal to
eight, our approximation is at most off by 10 to the power of negative six.

But remember, the question wants us
to find the least number of terms which has this property. We know that eight works, but we
need to check whether seven works. To do this, letβs start by
calculating the value of π. We can do this since weβre
calculating the infinite sum of a geometric series where the absolute value of the
ratio of successive terms is less than one. This infinite sum is equal to π
divided by one minus π. So, π is equal to one-fifth
divided by one minus negative one-fifth, which we can calculate to give us
one-sixth.

We could calculate the value of π
seven by using the formula for a finite sum of a geometric series. However, we can also just use a
calculator, giving us approximately 0.166665. If we then calculate the difference
between π and our seventh partial sum, we see itβs approximately 1.6 times 10 to
the power of negative six. And we see that this is bigger than
10 to the power of negative six. So, this means that π equals eight
mustβve been the least number of terms which guaranteed this level of accuracy. And we can then calculate π eight
to six decimal places to be 0.166666.

So, weβve now seen how to
approximate the infinite sum of an alternating series up to any level of accuracy
that we want by using a partial sum. And we can even find the least
number of terms needed for this level of accuracy. So, the key things weβve shown in
this video is if we have a decreasing and positive sequence π π whose limit as π
approaches β is equal to zero. Then, by the alternating series
test, the sum from π equals one to β of negative one to the power of π plus one
times π π is convergent β weβll call this value π. Then, we can approximate this
infinite sum by using a partial sum.

In fact, the absolute value of π
minus the πth partial sum is less than or equal to π π plus one. What this means is instead of
calculating the infinite sum, we can just calculate the sum for a finite number of
terms and then this will be accurate up to an error of most π π plus one. And if this is accurate up to an
error of most π π plus one, we could then use this to find the least number of
terms we would need for our partial sum to be accurate to a certain level of
accuracy. We find the sufficient number of
terms by using our bound, and then we remove terms until weβre no longer within the
required level of accuracy.