Calculus- Early Transcendentals, 2021a

352 Sequences and Series
Example 9.31
Approximate ∑1/n 2 to within 0.01.
N
Solution. Referring to Figure 9.2, if we approximate the sum by ∑ 1/n 2 , the size of the error we make is
n=1
the total area of the remaining rectangles, all of which lie under the curve 1/x 2 from x = N to infinity. So
we know the true value of the series is larger than the approximation, and no bigger than the approximation
plus the area under the curve from N to infinity. Roughly, then, we need to find N so that
∫ ∞
N
1
dx < 1/100.
x2 We can compute the integral:
∫ ∞ 1
N x 2 dx = 1 N ,
so if we choose N = 100 the error will be less than 0.01. Adding up the first 100 terms gives approximately
1.634983900. In fact, we can do a bit better. Since we know that the correct value is between our
approximation and our approximation plus the error (not minus), we can cut our error bound in half by
taking the value midway between these two values. If we take N = 50, we get a sum of 1.6251327 with an
error of at most 0.02, so the correct value is between 1.6251327 and 1.6451327, and therefore the value
halfway between these, 1.6351327, is within 0.01 of the correct value. We have mentioned that the true
value of this series can be shown to be π 2 /6 ≈ 1.644934068 which is 0.0098 more than our approximation,
and so (just barely) within the required error. Frequently approximations will be even better than the
“guaranteed” accuracy, but not always, as this example demonstrates.
♣
Exercises for 9.3
Determine whether each series converges or diverges.
Exercise 9.3.1
Exercise 9.3.2
Exercise 9.3.3
Exercise 9.3.4
∞
∑
n=1
∞
∑
n=1
1
n π/4
n
n 2 + 1
∞
lnn
∑
n=1
n 2
∞
∑
n=1
1
n 2 + 1
Exercise 9.3.5
Exercise 9.3.6
Exercise 9.3.7
Exercise 9.3.8
∞
∑
n=1
∞
∑
n=1
∞
∑
n=2
∞
∑
n=2
1
e n
n
e n
1
nlnn
1
n(lnn) 2
Exercise 9.3.9 Find an N so that
∞
∑
n=1
1
n 4 is between N
∑
n=1
1
n 4 and N
∑
n=1
1
n 4 + 0.005.