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Chap11_Sec3

# Chap11_Sec3 - 11 INFINITE SEQUENCES AND SERIES INFINITE...

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11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES

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INFINITE SEQUENCES AND SERIES 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 ∑ 1/[ n ( n +1)]. This is because, in each of these cases, we can find a simple formula for the n th partial sum s n . Nevertheless, usually, it isn’t easy to compute lim n n s →∞
INFINITE SEQUENCES AND SERIES So, in the next few sections, we develop several tests that help us determine whether a series is convergent or divergent without explicitly finding its sum. In some cases, however, our methods will enable us to find good estimates of the sum.

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INFINITE SEQUENCES AND SERIES Our first test involves improper integrals.
11.3 The Integral Test and Estimates of Sums In this section, we will learn how to: Find the convergence or divergence of a series and estimate its sum. INFINITE SEQUENCES AND SERIES

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We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers: There’s no simple formula for the sum s n of the first n terms. INTEGRAL TEST 2 2 2 2 2 2 1 1 1 1 1 1 1 ... 1 2 3 4 5 n n = = + + + + +
INTEGRAL TEST However, the computer-generated values given here suggest that the partial sums are approaching near 1.64 as n ∞. So, it looks as if the series is convergent. We can confirm this impression with a geometric argument.

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INTEGRAL TEST This figure shows the curve y = 1/ x 2 and rectangles that lie below the curve.
INTEGRAL TEST The base of each rectangle is an interval of length 1. The height is equal to the value of the function y = 1/ x 2 at the right endpoint of the interval.

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INTEGRAL TEST Thus, the sum of the areas of the rectangles is: 2 2 2 2 2 2 1 1 1 1 1 1 1 ... 1 2 3 4 5 n n = + + + + + =
If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y = 1/ x 2 for x ≥ 1, which is the value of the integral In Section 7.8, we discovered that this improper integral is convergent and has value 1. INTEGRAL TEST 2 1 (1/ ) x dx

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So, the image shows that all the partial sums are less than Therefore, the partial sums are bounded. INTEGRAL TEST 2 2 1 1 1 2 1 dx x + =
We also know that the partial sums are increasing (as all the terms are positive). Thus, the partial sums converge (by the Monotonic Sequence Theorem). So, the series is convergent. INTEGRAL TEST

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The sum of the series (the limit of the partial sums) is also less than 2: INTEGRAL TEST 2 2 2 2 2 1 1 1 1 1 1 ... 2 1 2 3 4 n n = = + + + + <
by the mathematician Leonhard Euler (1707–1783) to be π 2 /6. However, the proof of this fact is quite difficult.

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Chap11_Sec3 - 11 INFINITE SEQUENCES AND SERIES INFINITE...

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