Video: Finding the Maximum Error Bound When Approximating a Series by Summing Its First 20 Terms

Find the maximum error bound when approximating the series βˆ‘_(𝑛 = 1) ^(∞) ((βˆ’1)^(𝑛)) √((3𝑛 + 7)/(𝑛² + 1) by summing the first 20 terms. Round your answer to 5 decimal places.

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Video Transcript

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.

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