Video: Remainder of an Alternating Series

In this video, we will learn how to find the maximum error when approximating an alternating series by a finite term of the series.

17:51

Video Transcript

Remainder of an Alternating Series

In this video, we’re 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? Let’s start by discussing the types of alternating series we’re going to approximate.

Let π‘Ž 𝑛 be a positive and decreasing sequence where the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero. We chose these properties for π‘Ž 𝑛 so that it satisfies the alternating series test. This means the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times π‘Ž 𝑛 converges. Let’s call this value 𝑆. 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 𝑆. This means we can approximate 𝑆 by using the 𝑛th partial sum. We just take more and more terms to get a more and more accurate representation of 𝑆.

But just how accurate will this sum be? Let’s start by rewriting our infinite alternating series in terms of the 𝑛th partial sum. Our infinite alternating series will be equal to the π‘˜th partial sum plus all of the remaining terms. So, we have that our infinite series is equal to the π‘˜th partial sum plus the remainder after π‘˜ terms. We will call this 𝑅 π‘˜. And we want to know how close our estimate of the partial sum is to the actual value of 𝑆. If we have that 𝑆 is equal to the π‘˜th partial sum plus the remainder after π‘˜ terms, then by subtracting the π‘˜th partial sum from both sides of this equation, we have that 𝑆 minus 𝑆 π‘˜ is equal to 𝑅 π‘˜. In other words, the difference between the actual value of our series and our approximation is just equal to 𝑅 π‘˜. So, 𝑅 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 𝑆 minus the power series for the π‘˜th partial sum is equal to our remainder terms. So, to estimate how accurate our approximation is, we just need to estimate the value of 𝑅 π‘˜. To find an approximation of 𝑅 π‘˜, let’s 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‘Ž 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? Let’s compare the values of π‘Ž π‘˜ plus three and negative π‘Ž π‘˜ plus two. We know that π‘Ž π‘˜ plus three is smaller than π‘Ž π‘˜ plus two because our sequence is decreasing. This is the same as saying that negative π‘Ž π‘˜ plus two plus π‘Ž π‘˜ plus three is less than zero. What we’ve 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‘Ž π‘˜ plus five is smaller than π‘Ž 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‘Ž 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‘Ž π‘˜ plus one. We could then try pairing off the same terms we did before. However, this time we see that π‘Ž π‘˜ plus two is bigger than π‘Ž π‘˜ plus three. So, π‘Ž π‘˜ plus two minus π‘Ž π‘˜ plus three must be positive. In fact, this is true for all of these pairs. We’re subtracting a smaller positive number from a bigger positive number. And we see that the leading term of this series, negative π‘Ž 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‘Ž 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, we’ve shown that 𝑅 π‘˜ is less than or equal to π‘Ž π‘˜ plus one. And when π‘˜ is odd, we’ve shown that 𝑅 π‘˜ is greater than or equal to negative π‘Ž π‘˜ plus one. To combine these into one bound, we first want to show 𝑅 π‘˜ is positive when π‘˜ is even. To do this, let’s go back to our series expansion when π‘˜ was even, except this time we’re 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, 𝑅 π‘˜ is the sum of positive numbers, so it’s positive. Since π‘Ž 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 𝑅 π‘˜ is less than or equal to the first neglected term π‘Ž π‘˜ plus one.

We can do the same when π‘˜ is odd. We need to show that 𝑅 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, 𝑅 π‘˜ is negative. It’s 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‘Ž π‘˜ plus one is negative, and 𝑅 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‘Ž π‘˜ plus one. It’s just π‘Ž 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. Let’s see how we can use this to approximate an alternating series.

Find the maximum error bound when approximating the series the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power multiplied by the square root of three 𝑛 plus seven divided by 𝑛 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. That’s the same as taking the 20th partial sum. And the maximum possible error bound would be a bound on the absolute value of 𝑆 minus our 20th partial sum, where 𝑆 would be the value that our series converges to.

To help us find this bound, we know that if π‘Ž 𝑛 is a positive and decreasing sequence where the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero, then, by the alternating series test, the alternating series the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power times π‘Ž 𝑛 converges β€” we’ll call this value 𝑆.

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

Let’s start by showing that π‘Ž 𝑛 is positive. We know that 𝑛 is greater than or equal to one. So, three 𝑛 plus seven is positive, and then 𝑛 squared plus one is positive. And we’re taking the positive square root. So, our sequence π‘Ž 𝑛 is positive for all values of 𝑛 since we’re just taking the positive square root of a positive number.

To check that the sequence is decreasing, we’ll set 𝑓 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, we’ll start by setting 𝑒 equal to three π‘₯ plus seven divided by π‘₯ squared plus one and then using the chain rule. Since 𝑓 is a function of 𝑒 and 𝑒 is a function of π‘₯, by the chain rule we have 𝑓 prime of π‘₯ is equal to the derivative of the square root of 𝑒 with respect to 𝑒, which we can evaluate by using the power rule for differentiation. This gives us a half multiplied by 𝑒 to the power of negative a half. And then, we need to multiply this by d𝑒 by dπ‘₯.

Remember, we only need to calculate whether the slope is negative or positive. Since we’re only interested in the values of π‘₯ where π‘₯ is greater than or equal to one, we can see that 𝑒 as positive; it’s the quotient of two positive numbers. So, a half is positive, and 𝑒 to the power of negative a half is also positive. It’s 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 d𝑒 by dπ‘₯. To find d𝑒 by dπ‘₯, we’ll use the quotient rule. We’ll set 𝑣 equal to the numerator, three π‘₯ plus seven, and 𝑀 equal to the denominator, π‘₯ squared plus one. This gives us 𝑣 prime is three and 𝑀 prime is two π‘₯.

The quotient rule then tells us d𝑒 by dπ‘₯ is equal to 𝑣 prime 𝑀 minus 𝑣𝑀 prime all divided by 𝑀 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, d𝑒 by the dπ‘₯ is negative for these values of π‘₯. This means that our slope, 𝑓 prime of π‘₯, is a positive number multiplied by a negative number, which means 𝑓 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 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero. We see that π‘Ž 𝑛 is the square root of a fraction. We’ll divide both the numerator and the denominator of this fraction by the highest power of 𝑛 which appears in the fraction. That’s 𝑛 squared. Dividing through by 𝑛 squared gives us the limit as 𝑛 approaches ∞ of the square root of three over 𝑛 plus seven over 𝑛 squared all divided by one plus one over 𝑛 squared.

We can see as 𝑛 is approaching ∞, our numerator of three over 𝑛 plus seven over 𝑛 squared is approaching zero. And we can see the one over 𝑛 squared term in our denominator is also approaching zero. However, the term one remains constant. So, this fraction is approaching zero divided by one; it’s 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, we’ve shown as 𝑛 approaches ∞, π‘Ž 𝑛 approaches zero. This means the maximum error bound when summing the first 20 terms of our series is just π‘Ž 21. So, if we were to approximate this series by summing the first 20 terms, our error would be at most π‘Ž 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, we’ve now seen how to use this to find the maximum possible error. Let’s 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 𝑆 𝑛 for the least 𝑛 terms which guarantees that the sum of the first 𝑛 terms of the alternating series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one divided by five to the 𝑛th 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 𝑛, 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‘Ž equal to one-fifth and ratio of successive terms, π‘Ÿ, 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 we’re within 10 to the power of negative six of the infinite sum. And this would work.

However, we don’t know how many terms we would need to add. This could take hundreds and hundreds of terms for all we know. Instead, we’re going to try and approximate how many terms we would need. To help us estimate the value of 𝑛 we need, we recall the following fact about alternating series. If π‘Ž 𝑛 is a positive and decreasing sequence whose limit as 𝑛 approaches ∞ is equal to zero, then the alternating series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times π‘Ž 𝑛 is convergent by the alternating series test β€” we’ll call this equal to 𝑆. Then, we know we can bound the absolute value of 𝑆 minus the 𝑛th partial sum by the first term we left out. It’s less than or equal to π‘Ž 𝑛 plus one. We will set π‘Ž 𝑛 equal to one divided by five to the 𝑛th power.

We see if we set π‘Ž equal to one-fifth and π‘Ÿ also equal to one-fifth, then we can see that our sequence π‘Ž 𝑛 is also a geometric sequence. This actually then tells us all the information we need to know. One-fifth to the 𝑛th power is always positive. And the absolute value of one-fifth is less than one, so it’s decreasing. Finally, we know the limit as 𝑛 approaches ∞ of one divided by five to the 𝑛th power is equal to zero. So, we can use this to approximate the difference between the 𝑛th partial sum and the actual value of our infinite series 𝑆. The absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to one divided by five to the power of 𝑛 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 𝑛 such that one divided by five to the power of 𝑛 plus one was less than or equal to 10 to the power of negative six, then, for this particular value of 𝑛, the absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to 10 to the power of negative six. So, this value of 𝑛 is sufficient. However, it’s not necessarily the lowest value of 𝑛. So, let’s find the sufficient value of 𝑛. We want one divided by five to the power of 𝑛 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 𝑛 is greater than or equal to 7.6, which is the same as saying 𝑛 is greater than or equal to eight. So, what we have done is we have shown the absolute value of 𝑆 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 𝑛 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, let’s start by calculating the value of 𝑆. We can do this since we’re 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‘Ž divided by one minus π‘Ÿ. So, 𝑆 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 𝑆 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 𝑆 and our seventh partial sum, we see it’s 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 𝑛 equals eight must’ve been the least number of terms which guaranteed this level of accuracy. And we can then calculate 𝑆 eight to six decimal places to be 0.166666.

So, we’ve 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 we’ve shown in this video is if we have a decreasing and positive sequence π‘Ž 𝑛 whose limit as 𝑛 approaches ∞ is equal to zero. Then, by the alternating series test, the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times π‘Ž 𝑛 is convergent β€” we’ll call this value 𝑆. Then, we can approximate this infinite sum by using a partial sum.

In fact, the absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to π‘Ž 𝑛 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‘Ž 𝑛 plus one. And if this is accurate up to an error of most π‘Ž 𝑛 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 we’re no longer within the required level of accuracy.

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