James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 11.3 The Integral Test and Estimates of Sums 725
y
y=ƒ
a a£ a¢ a∞
0 1 2 3 4 5 . . . n x
FIGURE 5
y
y=ƒ
a n
Proof of the Integral Test
We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and
2 for the series o 1yn 2 and o 1ysn . For the general series o a n , look at Figures 5 and 6.
The area of the first shaded rectangle in Figure 5 is the value of f at the right endpoint of
f1, 2g, that is, f s2d − a 2 . So, comparing the areas of the shaded rectangles with the area
under y − f sxd from 1 to n, we see that
4
a 2 1 a 3 1 ∙ ∙ ∙ 1 a n < y n
f sxd dx
(Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6
shows that
1
5
y n
f sxd dx < a 1 1 a 2 1 ∙ ∙ ∙ 1 a n21
1
a n-1
a¡ a a£ a¢
0 1 2 3 4 5 . . . n x
FIGURE 6
(i) If y`
f sxd dx is convergent, then (4) gives
1
since f sxd > 0. Therefore
o n
a i < y n
f sxd dx < y`
f sxd dx
i−2 1
1
s n − a 1 1 o n
a i < a 1 1 y`
f sxd dx − M, say
i−2
Since s n < M for all n, the sequence hs n j is bounded above. Also
1
s n11 − s n 1 a n11 > s n
since a n11 − f sn 1 1d > 0. Thus hs n j is an increasing bounded sequence and so it is
con vergent by the Monotonic Sequence Theorem (11.1.12). This means that o a n is
convergent.
(ii) If y`
f sxd dx is divergent, then 1 yn f sxd dx l ` as n l ` because f sxd > 0. But
1
(5) gives
y n
n21
f sxd dx < o a i − s n21
1
i−1
and so s n21 l `. This implies that s n l ` and so o a n diverges.
n
1. Draw a picture to show that
1
ò
n−2 n , 1
y`
1.3 1.3
dx
1 x
What can you conclude about the series?
2. Suppose f is a continuous positive decreasing function for
x > 1 and a n − f snd. By drawing a picture, rank the following
three quantities in increasing order:
y 6
f sxd dx o 5
a i o 6
a i
1
i−1
i−2
3–8 Use the Integral Test to determine whether the series is
convergent or divergent.
3. ò n 23
n−1
2
5. ò
n−1 5n 2 1
n
7. ò
n−1 n 2 1 1
4. ò n 20.3
n−1
1
6. ò
n−1 s3n 2 1d 4
8. ò n 2 e 2n3
n−1
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