### Video Transcript

The series the sum from π equals
one to β of negative one to the power of π plus one times seven divided by the
square root of π can be approximated by summing the first seven terms. Find the maximum error bound of
this approximation.

The question tells us we can
approximate this infinite series by summing the first seven terms of this
series. In other words, weβre told that
this infinite series converges to some value. Letβs call this value π . And we can approximate the value of
π by summing the first seven terms. In other words, π is approximately
equal to the seventh partial sum of our series. The question wants to know the
maximum error bound of this approximation. The error in our approximation will
be the absolute value of π minus the seventh partial sum. And we want to find a bound on this
value.

We can see the series given to us in
the question appears to be an alternating series. And the bound weβre asked to find
looks very similar to one we know about certain alternating series. We recall if π π is a positive
and decreasing sequence where the limit as π approaches β of π π is equal to
zero, then the sum from π equals one to β of negative one to the power of ~~π
π plus one~~ [π plus one] times π π satisfies the alternating series
test. So this series converges and is
equal to some value we will call π . Then we can approximate the value
of this series π by using the πth partial sum. In particular, the absolute value
of π minus the πth partial sum is less than or equal to π π plus one.

And we can see the concluding
statement is very similar to what we are asked to find. In fact, the error of our
approximation is allowed to be equal to the maximum error bound since we just want
to know how big our error can possibly be. So to try and use this, weβll set
our sequence π π to be equal to seven divided by the square root of π. And if we can show that all of the
prerequisites are true by setting π equal to seven, we can calculate the maximum
error bound.

Letβs start by showing that π π is
a positive sequence. We see that π π is the quotient
of seven and the square root of π. We know that seven is a positive
number and the positive square root of π is also a positive number. So π π is always the quotient of
two positive numbers. Hence itβs always positive. Next, we need to show that π π is
a decreasing sequence. We might be tempted to try doing
this by setting π of π₯ equal to seven divided by the square root of π₯ and then
checking the slope of our function π of π₯. However, in this case, itβs easier
to show that our sequence is decreasing directly. We want to compare π π with π π
plus one.

We see that π π is seven divided
by the square root of π and that π π plus one is seven divided by the square root
of π plus one. Since seven, the square root of π,
and the square root of π plus one are all positive numbers and we know the square
root of π plus one is bigger than the square root of π. Weβre dividing by a larger positive
number in π π plus one. So π π plus one is smaller than
π π. And this is the same as saying that
our sequence π π is decreasing.

Finally, we need to show that the
limit as π approaches β of π π is equal to zero. And again, we can show this
directly. The limit as π approaches β of π
π is equal to the limit as π approaches β of seven divided by the square root of
π. Seven is a constant, so we can take
this outside of our limit to get seven times the limit as π approaches β of one
divided by the square root of π. And we know that this limit is
equal to zero.

So weβve now shown that all three
prerequisites are true. This means we can now bound the
error in our approximation. Since the question wants us to
approximate the value of π by using the first seven terms, weβll set π equal to
zero. And this means the absolute value
of π minus π seven is less than or equal to π eight. So π eight is the maximum possible
error bound of this approximation. We substitute π is equal to eight
into our sequence π π to get that π eight is equal to seven divided by the square
root of eight. And we could leave our answer like
this. However, we can simplify this
slightly. We know that eight is equal to four
times two. So the square root of eight is
equal to the square root of four times two, which by our laws of exponents is two
root two.

This gives us seven divided by two
root two. We can simplify this further by
multiplying the numerator and the denominator by the square root of two. In our numerator, we get seven root
two. And in our denominator, we get
~~seven~~ [two] root two times root two which simplifies to give us
four. So weβve shown if we approximate
the series the sum from π equals one to β of negative one to the power of π plus
one times seven divided by the square root of π by summing the first seven terms of
this series. Then the maximum error bound of
this approximation is seven root two divided by four.