Video: Finding the Maximum Error Bound of the Approximation of an Infinite Series

The series βˆ‘_(𝑛 = 1)^(∞) ((βˆ’1)^(𝑛 + 1) (7/√(𝑛))) can be approximated by summing the first 7 terms. Find the maximum error bound of this approximation.

04:30

Video Transcript

The series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times seven divided by the square root of 𝑛 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, we’re told that this infinite series converges to some value. Let’s call this value 𝑠. And we can approximate the value of 𝑠 by summing the first seven terms. In other words, 𝑠 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 𝑠 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 we’re asked to find looks very similar to one we know about certain alternating series. We recall 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 the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times π‘Ž 𝑛 satisfies the alternating series test. So this series converges and is equal to some value we will call 𝑠. Then we can approximate the value of this series 𝑠 by using the 𝑛th partial sum. In particular, the absolute value of 𝑠 minus the 𝑛th partial sum is less than or equal to π‘Ž 𝑛 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, we’ll set our sequence π‘Ž 𝑛 to be equal to seven divided by the square root of 𝑛. And if we can show that all of the prerequisites are true by setting 𝑛 equal to seven, we can calculate the maximum error bound.

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

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

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

So we’ve 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 𝑠 by using the first seven terms, we’ll set 𝑛 equal to zero. And this means the absolute value of 𝑠 minus 𝑠 seven is less than or equal to π‘Ž eight. So π‘Ž eight is the maximum possible error bound of this approximation. We substitute 𝑛 is equal to eight into our sequence π‘Ž 𝑛 to get that π‘Ž 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 two root two times root two which simplifies to give us four. So we’ve shown if we approximate the series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times seven divided by the square root of 𝑛 by summing the first seven terms of this series. Then the maximum error bound of this approximation is seven root two divided by four.

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