### Video Transcript

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.