### Video Transcript

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.