CalculusVolume2-OP-98

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5.14
Testing for Convergence of p-series
For each of the following series, determine whether it converges or diverges.
a. ∑
n = 1
∞
1
n4
b. ∑
n = 1
∞
1
n2/3
Solution
a. This is a p-series with p = 4 > 1, so the series converges.
b. Since p = 2/3 < 1, the series diverges.
Does the series ∑
n = 1
∞
1
n5/4 converge or diverge?
Estimating the Value of a Series
Suppose we know that a series ∑
n = 1
∞
an converges and we want to estimate the sum of that series. Certainly we can
approximate that sum using any finite sum ∑
n = 1
N
an where N is any positive integer. The question we address here is, for
a convergent series ∑
n = 1
∞
an, how good is the approximation ∑
n = 1
N
an ? More specifically, if we let
RN = ∑
n = 1
∞
an − ∑
n = 1
N
an
be the remainder when the sum of an infinite series is approximated by the Nth partial sum, how large is RN ? For some
types of series, we are able to use the ideas from the integral test to estimate RN.
Theorem 5.10: Remainder Estimate from the Integral Test
Suppose ∑
n = 1
∞
an is a convergent series with positive terms. Suppose there exists a function f satisfying the following
three conditions:
i. f is continuous,
ii. f is decreasing, and
iii. f (n) = an for all integers n ≥ 1.
Let SN be the Nth partial sum of ∑
n = 1
∞
an. For all positive integers N,
478 Chapter 5 | Sequences and Series
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SN + ∫
N + 1
∞
f (x)dx < ∑
n = 1
∞
an < SN + ∫
N
∞
f (x)dx.
In other words, the remainder RN = ∑
n = 1
∞
an − SN = ∑
n = N + 1
∞
an satisfies the following estimate:
(5.10)∫
N + 1
∞
f (x)dx < RN < ∫
N
∞
f (x)dx.
This is known as the remainder estimate.
We illustrate Remainder Estimate from the Integral Test in Figure 5.15. In particular, by representing the remainder
RN = aN + 1 + aN + 2 + aN + 3 + ⋯ as the sum of areas of rectangles, we see that the area of those rectangles is bounded
above by ∫
N
∞
f (x)dx and bounded below by ∫
N + 1
∞
f (x)dx. In other words,
RN = aN + 1 + aN + 2 + aN + 3 + ⋯ > ∫
N + 1
∞
f (x)dx
and
RN = aN + 1 + aN + 2 + aN + 3 + ⋯ < ∫
N
∞
f (x)dx.
We conclude that
∫
N + 1
∞
f (x)dx < RN < ∫
N
∞
f (x)dx.
Since
∑
n = 1
∞
an = SN + RN,
where SN is the Nth partial sum, we conclude that
SN + ∫
N + 1
∞
f (x)dx < ∑
n = 1
∞
an < SN + ∫
N
∞
f (x)dx.
Chapter 5 | Sequences and Series 479
Figure 5.15 Given a continuous, positive, decreasing function f and a sequence of positive
terms an such that an = f (n) for all positive integers n, (a) the areas
aN + 1 + aN + 2 + aN + 3 + ⋯ < ∫
N
∞
f (x)dx, or (b) the areas
aN + 1 + aN + 2 + aN + 3 + ⋯ > ∫
N + 1
∞
f (x)dx. Therefore, the integral is either an
overestimate or an underestimate of the error.
Example 5.16
Estimating the Value of a Series
Consider the series ∑
n = 1
∞
1/n3.
a. Calculate S10 = ∑
n = 1
10
1/n3 and estimate the error.
b. Determine the least value of N necessary such that SN will estimate ∑
n = 1
∞
1/n3 to within 0.001.
Solution
a. Using a calculating utility, we have
S10 = 1 + 1
23 + 1
33 + 1
43 + ⋯ + 1
103 ≈ 1.19753.
By the remainder estimate, we know
RN < ∫
N
∞
1
x3dx.
We have
∫
10
∞
1
x3dx = lim
b → ∞
∫
10
b
1
x3dx = lim
b → ∞
⎡
⎣− 1
2x2
⎤
⎦
N
b
= lim
b → ∞
⎡
⎣− 1
2b2 + 1
2N 2
⎤
⎦ = 1
2N 2.
480 Chapter 5 | Sequences and Series
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5.15
Therefore, the error is R10 < 1/2(10)2 = 0.005.
b. Find N such that RN < 0.001. In part a. we showed that RN < 1/2N 2. Therefore, the remainder
RN < 0.001 as long as 1/2N 2 < 0.001. That is, we need 2N 2 > 1000. Solving this inequality for
N, we see that we need N > 22.36. To ensure that the remainder is within the desired amount, we need
to round up to the nearest integer. Therefore, the minimum necessary value is N = 23.
For ∑
n = 1
∞
1
n4, calculate S5 and estimate the error R5.
Chapter 5 | Sequences and Series 481
5.3 EXERCISES
For each of the following sequences, if the divergence test
applies, either state that limn → ∞an does not exist or find
limn → ∞an. If the divergence test does not apply, state why.
138. an = n
n + 2
139. an = n
5n2 − 3
140. an = n
3n2 + 2n + 1
141. an = (2n + 1)(n − 1)
(n + 1)2
142. an = (2n + 1)2n
⎛
⎝3n2 + 1⎞
⎠
n
143. an = 2n
3n/2
144. an = 2n + 3n
10n/2
145. an = e−2/n
146. an = cosn
147. an = tann
148. an = 1 − cos2 (1/n)
sin2 (2/n)
149. an = ⎛
⎝1 − 1
n
⎞
⎠
2n
150. an = lnn
n
151. an = (lnn)2
n
State whether the given p -series converges.
152. ∑
n = 1
∞
1
n
153. ∑
n = 1
∞
1
n n
154. ∑
n = 1
∞
1
n23
155. ∑
n = 1
∞
1
n43
156. ∑
n = 1
∞
ne
nπ
157. ∑
n = 1
∞
nπ
n2e
Use the integral test to determine whether the following
sums converge.
158. ∑
n = 1
∞
1
n + 5
159. ∑
n = 1
∞
1
n + 53
160. ∑
n = 2
∞
1
n lnn
161. ∑
n = 1
∞
n
1 + n2
162. ∑
n = 1
∞
en
1 + e2n
163. ∑
n = 1
∞
2n
1 + n4
164. ∑
n = 2
∞
1
n ln2 n
Express the following sums as p -series and determine
whether each converges.
165. ∑
n = 1
∞
2−lnn (Hint: 2−lnn = 1/nln2 .)
166. ∑
n = 1
∞
3−lnn (Hint: 3−lnn = 1/nln3 .)
167. ∑
n = 1
∞
n2−2lnn
482 Chapter 5 | Sequences and Series
This OpenStax book is available for free at http://cnx.org/content/col11965/1.2