39–40. {Use of Tech} Lower and upper bounds of a series
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Question

39–40. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, use Theorem 10.13 to complete the following.

c. Find lower and upper bounds (Lₙ and Uₙ, respectively) for the exact value of the series.

39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the lower bound of the exact value of the convergence series, the sum of K equal to 1 to infinity of 1 divided by K4 when N is equal to 2. Round the answer to 4 decimal places. So first we want to define our partial sums and our remainders. So we call that our partial sum, which will denote as SN, it's going to be equal to the sum. Of K equal to 1 ton of 1 divided by K 4th. And our remainder, our sub N, is going to be equal to the sum of K equal to N + 1 to infinity of 1 divided by K4th. Now, since 1 divided by K 4th is a decreasing quantity as K increases, we can actually use the integral test to estimate the bounds for our remainder. And so we recall for the integral test that we have the integral from N + 1 to infinity of F of XDX. is less than or equal torn, which is less than or equal to the integral from N to infinity of F of X D X, where in our case F of X is going to be equal to 1 divided by x to the 4th, and we're given the problem that we want n to be equal to 2. However, we're looking for the lower bound of the exact value, not the remainder. However, if I look at SN plus RN, That's just going to be equal to our entire series, which is going to be the sum of K equal to 1, to infinity of 1 divided by K 4th, which would give us the exact value of our convergence series. So, what we're going to do is add S of N to each term in our inequality here. So, that means that S of N. Plus the integral from N + 1 to infinity of F of X. D X is less than or equal to sn plus RN, which is just our total sum, which is going to be the sum of K equal to 1 to infinity of 1 divided by k 4th, which is less than or equal to S N plus the interval from N to infinity of F of X D X. So here, now we have bounds on our exact value. And we're interested in the lower bound, which is going to be the SN plus the integral of N plus the integral from N + 1 to infinity of F of X D X. And so we're going to look at each of those terms separately. And we're looking for the case when N is equal to 2, so we need to find S2, which is just going to be equal to the first two terms in our sum, which is going to be 1 divided by 1 to the 4th. Plus 1 divided by 24th. And when we add these two together, we see that this is just going to be equal to 17 divided by 16. Next we need to evaluate the interval, and so that is going to be the integral. From 3 to infinity of F of X, which is 1 divided by X to the fourth, DX. And this is the improper interval, so we're going to rewrite this as being equal to the limit, as B approaches infinity of the integral from 3 to B. Of 1 divided by X to the 4th DX. So, performing the integration, this is going to be equal to the limit, as B approaches infinity. Of the quantity of minus 1 divided by the quantity of 3 X cubed in quantity, and this needs to be evaluated from 3 to B. So sub between in the limits of integration, you see that this is going to be equal to the limit. As B approaches infinity. Of the quantity of minus 1 divided by the quantity of 3 be cued in quantity, plus 1 divided by the quantity of 3 multiplied by 3 cubed in quantity. Now, as B approaches infinity. 1 divided by the quantity of 3 multiplied by B cubed, and quantity is going to approach 0. So that means that this is going to be equal to 0 plus 1 divided by 81, which is just going to be equal to 1 divided by 81. So that means that if we look at our lower bound, For the exact value, we will have that S2 plus the integral from 3 to infinity of 1 divided by X to the 4th DX is going to be equal to. 17 divided by 16. Plus 1 divided by 81. And when we evaluate this expression, this is going to be approximately equal to. 1.0748. So here we've just rounded to 4 decimal places, and this is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

39–40. {Use of Tech} Lower and upper bounds of a series
For each convergent series and given value of n, use Theorem 10.13 to complete the following.

c. Find lower and upper bounds (Lₙ and Uₙ, respectively) for the exact value of the series.

39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Key Concepts

Convergent Series

Theorem 10.13 (Bounds for Series Sums)

Partial Sums and Remainder Estimation

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