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# 11.3

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## 11.3 The Integral Test and

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### The Integral Test and Estimates of Sums

In general, it is difficult to find the exact sum of a series.

We were able to accomplish this for geometric series and the series Σ 1/[n(n + 1)] because in each of those cases we could find a simple formula for the nth partial sum sn.

But usually it isn’t easy to discover such a formula.

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### The Integral Test and Estimates of Sums

We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers:

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### The Integral Test and Estimates of Sums

There’s no simple formula for the sum sn of the first n terms, but the computer-generated table of approximate values

given in the margin suggests that the partial sums are approaching a number near 1.64 as n → and

so it looks as if the series is convergent.

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### The Integral Test and Estimates of Sums

We can confirm this impression with a geometric argument.

Figure 1 shows the curve y = 1/x2 and rectangles that lie below the curve.

The base of each rectangle is an interval of length 1; the height is equal to the value of the function y = 1/x2 at the

Figure 1

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### The Integral Test and Estimates of Sums

So the sum of the areas of the rectangles is

If we exclude the first rectangle, the total area of the

remaining rectangles is smaller than the area under the curve y = 1/x2 for x ≥ 1, which is the value of the

integral

This improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than

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### The Integral Test and Estimates of Sums

Thus the partial sums are bounded. We also know that the partial sums are increasing (because all the terms are

positive). Therefore the partial sums converge (by the Mono tonic Sequence Theorem) and so the series is

convergent. The sum of the series (the limit of the partial sums) is also less than 2:

Now let’s look at the series

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### The Integral Test and Estimates of Sums

The table of values of sn suggests that the partial sums

aren’t approaching a finite number, so we suspect that the given series may be divergent.

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Again we use a picture for confirmation. Figure 2 shows the curve but this time we use rectangles whose tops lie above the curve.

The base of each rectangle is an interval of length 1. The height is equal to the value of the function at the left endpoint of the interval.

Figure 2

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### The Integral Test and Estimates of Sums

So the sum of the areas of all the rectangles is

This total area is greater than the area under the curve for x ≥ 1, which is equal to the integral

But we know that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite, that is, the series is

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### The Integral Test and Estimates of Sums

The same sort of geometric reasoning that we used for these two series can be used to prove the following test.

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### Example 1

Test the series for convergence or divergence.

Solution:

The function f(x) = 1/(x2 + 1) is continuous, positive, and decreasing on [1, ) so we use the Integral Test:

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### The Integral Test and Estimates of Sums

The series is called the p-series.

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### Example 3

(a) The series

is convergent because it is a p-series with p = 3 > 1.

(b) The series

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### Example 4

Determine whether the series converges or diverges.

Solution:

The function f(x) = (ln x)/x is positive and continuous for x > 1 because the logarithm function is continuous.

But it is not obvious whether or not f is decreasing, so we compute its derivative:

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### Example 4

Thus f'(x) < 0 when ln x > 1, that is, x > e. It follows that f is decreasing when x > e and so we can apply the Integral Test:

Since this improper integral is divergent, the series

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### Estimating the Sum of a Series

Suppose we have been able to use the Integral Test to show that a series Σ an is convergent and we now want to find an approximation to the sum s of the series.

Of course, any partial sum sn is an approximation to s because limn sn = s. But how good is such an

approximation? To find out, we need to estimate the size of the remainder

Rn = s – sn = an + 1 + an + 2 + an + 3 +

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### Estimating the Sum of a Series

We use the same notation and ideas as in the Integral Test, assuming that f is decreasing on [n, ). Comparing the

areas of the rectangles with the area under y = f(x) for x > n in Figure 3, we see that

Similarly, we see from Figure 4 that Figure 3

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### Estimating the Sum of a Series

So we have proved the following error estimate.

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### Example 5

(a) Approximate the sum of the series Σ 1/n3 by using the sum of the first 10 terms. Estimate the error involved in this approximation.

(b) How many terms are required to ensure that the sum is accurate to within 0.0005?

Solution:

In both parts (a) and (b) we need to know With

f(x) = 1/x3, which satisfies the conditions of the Integral Test, we have

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### Example 5 – Solution

(a) Approximating the sum of the series by the 10th partial sum, we have

According to the remainder estimate in (2), we have

cont’d

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### Example 5 – Solution

(b) Accuracy to within 0.0005 means that we have to find a value of n such that Rn ≤ 0.0005. Since

we want

Solving this inequality, we get

We need 32 terms to ensure accuracy to within 0.0005.

cont’d

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### Estimating the Sum of a Series

If we add sn to each side of the inequalities in , we get

because sn + Rn = s. The inequalities in (3) give a lower bound and an upper bound for s.

They provide a more accurate approximation to the sum of the series than the partial sum s does.

The following theorem states that (1 + x) k is equal to the sum of its Maclaurin

Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure

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