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C H A P T E R
Infinite Series, Power Series
1. THE GEOMETRIC SERIES
As a simple example of many of the ideas involved in series, we are going to consider the geometric series. You may recall that in a geometric progression we multiply each term by some fixed number to get the next term. For example, the sequences
(1.1a) 2 , 4 , 8 , 16 , 32 ,... , (1.1b) 1 , 23 , 49 , 278 , 1681 ,... , (1.1c) a, ar, ar^2 , ar^3 ,... ,
are geometric progressions. It is easy to think of examples of such progressions. Suppose the number of bacteria in a culture doubles every hour. Then the terms of (1.1a) represent the number by which the bacteria population has been multiplied after 1 hr, 2 hr, and so on. Or suppose a bouncing ball rises each time to 23 of the height of the previous bounce. Then (1.1b) would represent the heights of the successive bounces in yards if the ball is originally dropped from a height of 1 yd. In our first example it is clear that the bacteria population would increase with- out limit as time went on (mathematically, anyway; that is, assuming that nothing like lack of food prevented the assumed doubling each hour). In the second example, however, the height of bounce of the ball decreases with successive bounces, and we might ask for the total distance the ball goes. The ball falls a distance 1 yd, rises a distance 23 yd and falls a distance 23 yd, rises a distance 49 yd and falls a distance 4 9 yd, and so on. Thus it seems reasonable to write the following expression for the total distance the ball goes:
(1.2) 1 + 2 · 23 + 2 · 49 + 2 · 278 + · · · = 1 + 2
3 +^
4 9 +^
8 27 +^ · · ·^
where the three dots mean that the terms continue as they have started (each one being 23 the preceding one), and there is never a last term. Let us consider the expression in parentheses in (1.2), namely
2 Infinite Series, Power Series Chapter 1
This expression is an example of an infinite series, and we are asked to find its sum. Not all infinite series have sums; you can see that the series formed by adding the terms in (1.1a) does not have a finite sum. However, even when an infinite series does have a finite sum, we cannot find it by adding the terms because no matter how many we add there are always more. Thus we must find another method. (It is actually deeper than this; what we really have to do is to define what we mean by the sum of the series.) Let us first find the sum of n terms in (1.3). The formula (Problem 2) for the sum of n terms of the geometric progression (1.1c) is
(1.4) Sn =
a(1 − rn) 1 − r
Using (1.4) in (1.3), we find
(1.5) Sn =
)n
2 3 [1^ −^ (^
2 3 )
n] 1 − (^23)
[
)n] .
As n increases, ( 23 )n^ decreases and approaches zero. Then the sum of n terms approaches 2 as n increases, and we say that the sum of the series is 2. (This is really a definition: The sum of an infinite series is the limit of the sum of n terms as n → ∞.) Then from (1.2), the total distance traveled by the ball is 1 + 2 · 2 = 5. This is an answer to a mathematical problem. A physicist might well object that a bounce the size of an atom is nonsense! However, after a number of bounces, the remaining infinite number of small terms contribute very little to the final answer (see Problem 1). Thus it makes little difference (in our answer for the total distance) whether we insist that the ball rolls after a certain number of bounces or whether we include the entire series, and it is easier to find the sum of the series than to find the sum of, say, twenty terms. Series such as (1.3) whose terms form a geometric progression are called geo- metric series. We can write a geometric series in the form
(1.6) a + ar + ar^2 + · · · + arn−^1 + · · ·.
The sum of the geometric series (if it has one) is by definition
(1.7) S = lim n→∞ Sn,
where Sn is the sum of n terms of the series. By following the method of the exam- ple above, you can show (Problem 2) that a geometric series has a sum if and only if |r| < 1, and in this case the sum is
(1.8) S =
a 1 − r
4 Infinite Series, Power Series Chapter 1
- Suppose a large number of particles are bouncing back and forth between x = 0 and x = 1, except that at each endpoint some escape. Let r be the fraction reflected each time; then (1 − r) is the fraction escaping. Suppose the particles start at x = 0 heading toward x = 1; eventually all particles will escape. Write an infinite series for the fraction which escape at x = 1 and similarly for the fraction which escape at x = 0. Sum both the series. What is the largest fraction of the particles which can escape at x = 0? (Remember that r must be between 0 and 1.)
2. DEFINITIONS AND NOTATION
There are many other infinite series besides geometric series. Here are some exam- ples:
(2.1a) 12 + 2^2 + 3^2 + 4^2 + · · · , 1 2
(2.1b) + · · · ,
x − x^2 2
x^3 3
x^4 4
(2.1c) + · · ·.
In general, an infinite series means an expression of the form
(2.2) a 1 + a 2 + a 3 + · · · + an + · · · ,
where the an’s (one for each positive integer n) are numbers or functions given by some formula or rule. The three dots in each case mean that the series never ends. The terms continue according to the law of formation, which is supposed to be evident to you by the time you reach the three dots. If there is apt to be doubt about how the terms are formed, a general or nth term is written like this:
(2.3a) 12 + 2^2 + 3^2 + · · · + n^2 + · · · ,
x − x^2 +
x^3 2
(−1)n−^1 xn (n − 1)! (2.3b) + · · ·.
(The quantity n!, read n factorial, means, for integral n, the product of all integers from 1 to n; for example, 5! = 5 · 4 · 3 · 2 · 1 = 120. The quantity 0! is defined to be 1.) In (2.3a), it is easy to see without the general term that each term is just the square of the number of the term, that is, n^2. However, in (2.3b), if the formula for the general term were missing, you could probably make several reasonable guesses for the next term. To be sure of the law of formation, we must either know a good many more terms or have the formula for the general term. You should verify that the fourth term in (2.3b) is −x^4 /6. ∑ We can also write series in a shorter abbreviated form using a summation sign followed by the formula for the nth term. For example, (2.3a) would be written
(2.4) 12 + 2^2 + 3^2 + 4^2 + · · · =
∑^ ∞
n=
n^2
(read “the sum of n^2 from n = 1 to ∞”). The series (2.3b) would be written
x − x^2 +
x^3 2
x^4 6
∑^ ∞
n=
(−1)n−^1 xn (n − 1)!
Section 2 Definitions and Notation 5
For printing convenience, sums like (2.4) are often written
n=1 n
In Section 1, we have mentioned both sequences and series. The lists in (1.1) are sequences; a sequence is simply a set of quantities, one for each n. A series is an indicated sum of such quantities, as in (1.3) or (1.6). We will be interested in various sequences related to a series: for example, the sequence an of terms of the series, the sequence Sn of partial sums [see (1.5) and (4.5)], the sequence Rn [see (4.7)], and the sequence ρn [see (6.2)]. In all these examples, we want to find the limit of a sequence as n → ∞ (if the sequence has a limit). Although limits can be found by computer, many simple limits can be done faster by hand.
Example 1. Find the limit as n → ∞ of the sequence
(2n − 1)^4 +
1 + 9n^8 1 − n^3 − 7 n^4
We divide numerator and denominator by n^4 and take the limit as n → ∞. Then all terms go to zero except 24 +
Example 2. Find limn→∞ ln n^ n. By L’Hˆopital’s rule (see Section 15)
lim n→∞
ln n n
= lim n→∞
1 /n 1
Comment: Strictly speaking, we can’t differentiate a function of n if n is an integer, but we can consider f (x) = (ln x)/x, and the limit of the sequence is the same as the limit of f (x).
Example 3. Find limn→∞
n
) 1 /n
. We first find
ln
n
) 1 /n = −
n
ln n.
Then by Example 2, the limit of (ln n)/n is 0, so the original limit is e^0 = 1.
PROBLEMS, SECTION 2
In the following problems, find the limit of the given sequence as n → ∞.
n^2 + 5n^3 2 n^3 + 3
√ 4 + n^6
(n + 1)^2 √ 3 + 5n^2 + 4n^4
(−1)n
√ n + 1 n
2 n n^2
10 n n!
nn n!
- (1 + n^2 )^1 /^ ln^ n^ 8.^
(n!)^2 (2n)!
- n sin(1/n)
Section 4 Convergent and Divergent Series 7
is convergent as it stands, but can be made to have any sum you like by combining the terms in a different order! (See Section 8.) You can see from these examples how essential it is to know whether a series converges, and also to know how to apply algebra to series correctly. There are even cases in which some divergent series can be used (see Chapter 11), but in this chapter we shall be concerned with convergent series. Before we consider some tests for convergence, let us repeat the definition of convergence more carefully. Let us call the terms of the series an so that the series is
(4.4) a 1 + a 2 + a 3 + a 4 + · · · + an + · · ·.
Remember that the three dots mean that there is never a last term; the series goes on without end. Now consider the sums Sn that we obtain by adding more and more terms of the series. We define
S 1 = a 1 , S 2 = a 1 + a 2 , S 3 = a 1 + a 2 + a 3 , · · · Sn = a 1 + a 2 + a 3 + · · · + an.
Each Sn is called a partial sum; it is the sum of the first n terms of the series. We had an example of this for a geometric progression in (1.4). The letter n can be any integer; for each n, Sn stops with the nth term. (Since Sn is not an infinite series, there is no question of convergence for it.) As n increases, the partial sums may increase without any limit as in the series (2.1a). They may oscillate as in the series 1 − 2 + 3 − 4 + 5 − · · · (which has partial sums 1, − 1 , 2 , − 2 , 3 , · · · ) or they may have some more complicated behavior. One possibility is that the Sn’s may, after a while, not change very much any more; the an’s may become very small, and the Sn’s come closer and closer to some value S. We are particularly interested in this case in which the Sn’s approach a limiting value, say
lim n→∞ (4.6) Sn = S.
(It is understood that S is a finite number.) If this happens, we make the following definitions.
a. If the partial sums Sn of an infinite series tend to a limit S, the series is called convergent. Otherwise it is called divergent.
b. The limiting value S is called the sum of the series.
c. The difference Rn = S − Sn is called the remainder (or the remainder after n terms). From (4.6), we see that
lim n→∞ Rn = lim n→∞ (4.7) (S − Sn) = S − S = 0.
8 Infinite Series, Power Series Chapter 1
Example 1. We have already (Section 1) found Sn and S for a geometric series. From (1.8) and (1.4), we have for a geometric series, Rn = ar
n 1 −r which^ →^ 0 as^ n^ → ∞^ if^ |r|^ <^ 1.
Example 2. By partial fractions, we can write (^) n (^22) − 1 = (^) n^1 − 1 − (^) n+1^1. Let’s write out a number of terms of the series ∑^ ∞
2
n^2 − 1
∑^ ∞
2
n − 1
n + 1
∑^ ∞
1
n
n + 2
n − 2
n
n − 1
n + 1
n
n + 2
Note the cancellation of terms; this kind of series is called a telescoping series. Satisfy yourself that when we have added the nth term ( (^1) n − (^) n^1 +2 ), the only terms which have not cancelled are 1, 12 , (^) n−+1^1 , and (^) n−+2^1 , so we have
Sn =
n + 1
n + 2
, S =
, Rn =
n + 1
n + 2
Example 3. Another interesting series is
∑^ ∞
1
ln
n n + 1
∑^ ∞
1
[ln n − ln(n + 1)]
= ln 1 − ln 2 + ln 2 − ln 3 + ln 3 − ln 4 + · · · + ln n − ln(n + 1) · · ·.
Then Sn = − ln(n + 1) which → −∞ as n → ∞, so the series diverges. However, note that an = ln (^) nn+1 → ln 1 = 0 as n → ∞, so we see that even if the terms tend to zero, a series may diverge.
PROBLEMS, SECTION 4
For the following series, write formulas for the sequences an, Sn, and Rn, and find the limits of the sequences as n → ∞ (if the limits exist).
X^ ∞
1
1 2 n^
X^ ∞
0
1 5 n
1 − 1 2 + 1 4 − 1 8 + 1 16 · · ·
X^ ∞
1
e−n^ ln 3^ Hint: What is e−^ ln 3?
X^ ∞ 0
e^2 n^ ln sin(π/3)^ Hint: Simplify this.
X^ ∞
1
1 n(n + 1)
Hint: 1 n(n + 1)
=^1 n
− 1 n + 1
.
3 1 · 2 − 5 2 · 3
7 3 · 4 − 9 4 · 5
10 Infinite Series, Power Series Chapter 1
6. CONVERGENCE TESTS FOR SERIES OF POSITIVE TERMS;
ABSOLUTE CONVERGENCE
We are now going to consider four useful tests for series whose terms are all positive. If some of the terms of a series are negative, we may still want to consider the related series which we get by making all the terms positive; that is, we may consider the series whose terms are the absolute values of the terms of our original series. If this new series converges, we call the original series absolutely convergent. It can be proved that if a series converges absolutely, then it converges (Problem 7.9). This means that if the series of absolute values converges, the series is still convergent when you put back the original minus signs. (The sum is different, of course.) The following four tests may be used, then, either for testing series of positive terms, or for testing any series for absolute convergence.
A. The Comparison Test
This test has two parts, (a) and (b).
(a) Let m 1 + m 2 + m 3 + m 4 + · · · be a series of positive terms which you know converges. Then the series you are testing, namely a 1 + a 2 + a 3 + a 4 + · · · is absolutely convergent if |an| ≤ mn (that is, if the absolute value of each term of the a series is no larger than the corresponding term of the m series) for all n from some point on, say after the third term (or the millionth term). See the example and discussion below. (b) Let d 1 + d 2 + d 3 + d 4 + · · · be a series of positive terms which you know diverges. Then the series
|a 1 | + |a 2 | + |a 3 | + |a 4 | + · · ·
diverges if |an| ≥ dn for all n from some point on. Warning: Note carefully that neither |an| ≥ mn nor |an| ≤ dn tells us anything. That is, if a series has terms larger than those of a convergent series, it may still converge or it may diverge—we must test it further. Similarly, if a series has terms smaller than those of a divergent series, it may still diverge, or it may converge.
Example. Test
∑^ ∞
n=
n!
As a comparison series, we choose the geometric series
∑^ ∞
n=
2 n^
Notice that we do not care about the first few terms (or, in fact, any finite number of terms) in a series, because they can affect the sum of the series but not whether
Section 6 Convergence Tests for Series of Positive Terms; Absolute Convergence 11
it converges. When we ask whether a series converges or not, we are asking what happens as we add more and more terms for larger and larger n. Does the sum increase indefinitely, or does it approach a limit? What the first five or hundred or million terms are has no effect on whether the sum eventually increases indefinitely or approaches a limit. Consequently we frequently ignore some of the early terms in testing series for convergence. In our example, the terms of
n=1 1 /n! are smaller than the corresponding terms of
n=1 1 /^2
n (^) for all n > 3 (Problem 1). We know that the geometric series converges because its ratio is 12. Therefore
n=1 1 /n! converges also.
PROBLEMS, SECTION 6
- Show that n! > 2 n^ for all n > 3. Hint: Write out a few terms; then consider what you multiply by to go from, say, 5! to 6! and from 2^5 to 2^6.
- Prove that the harmonic series P∞ n=1 1 /n^ is divergent by comparing it with the series
1 +^1 2
„ 1 4 +^1 4
«
„ 1 8 +^1 8 +^1 8 +^1 8
«
„ 8 terms each equal to 1 16
«
which is 1 +^1 2
+^1 2
+^1 2
+^1 2
- Prove the convergence of
P∞ n=1 1 /n
(^2) by grouping terms somewhat as in Problem 2.
- Use the comparison test to prove the convergence of the following series: (a)
X^ ∞ n=
1 2 n^ + 3n^ (b)
X^ ∞ n=
1 n 2 n
- Test the following series for convergence using the comparison test. (a)
X^ ∞
n=
√^1 n Hint: Which is larger, n or
√ n? (b)
X^ ∞
n=
1 ln n
- There are 9 one-digit numbers (1 to 9), 90 two-digit numbers (10 to 99). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series 1 + 12 + 13 + · · · + 19 are all greater than 101 ; similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series ˆ (^1) 10 +^ 1 10 +^ · · ·^ (9 terms each =^ 1 10 )
˜
ˆ 90 terms each = 1001
˜
- · · · = 109 + 10090 + · · · = 109 + 109 + · · ·.
The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory m series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.
B. The Integral Test
We can use this test when the terms of the series are positive and not increasing, that is, when an+1 ≤ an. (Again remember that we can ignore any finite number of terms of the series; thus the test can still be used even if the condition an+1 ≤ an does not hold for a finite number of terms.) To apply the test we think of an as a
Section 6 Convergence Tests for Series of Positive Terms; Absolute Convergence 13
PROBLEMS, SECTION 6
Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals (see Problem 16).
X^ ∞
n=
1 n ln n
X^ ∞
n=
n n^2 + 4
X^ ∞
n=
1 n^2 − 4
X^ ∞
n=
en e^2 n^ + 9
X^ ∞
1
1 n(1 + ln n)^3 /^2
X^ ∞
1
n (n^2 + 1)^2
X^ ∞
1
n^2 n^3 + 1
X^ ∞
1
√^1 n^2 + 9
- Use the integral test to prove the following so-called p-series test. The series
X^ ∞
n=
1 np^ is
( convergent if p > 1 , divergent if p ≤ 1.
Caution: Do p = 1 separately.
- In testing
P 1 /n^2 for convergence, a student evaluates
R (^) ∞ 0 n
− (^2) dn = −n− (^1) |∞ 0 = 0 + ∞ = ∞ and concludes (erroneously) that the series diverges. What is wrong? Hint: Consider the area under the curve in a diagram such as Figure 6.1 or 6.2. This example shows the danger of using a lower limit in the integral test.
- Use the integral test to show that
P∞ n=0 e
−n^2 converges. Hint: Although you cannot evaluate the integral, you can show that it is finite (which is all that is necessary) by comparing it with
R (^) ∞ e−ndn.
C. The Ratio Test
The integral test depends on your being able to integrate andn; this is not always easy! We consider another test which will handle many cases in which we cannot evaluate the integral. Recall that in the geometric series each term could be obtained by multiplying the one before it by the ratio r, that is, an+1 = ran or an+1/an = r. For other series the ratio an+1/an is not constant but depends on n; let us call the absolute value of this ratio ρn. Let us also find the limit (if there is one) of the sequence ρn as n → ∞ and call this limit ρ. Thus we define ρn and ρ by the equations
ρn =
an+ an
ρ = lim n→∞ ρn.
If you recall that a geometric series converges if |r| < 1, it may seem plausible that a series with ρ < 1 should converge and this is true. This statement can be proved (Problem 30) by comparing the series to be tested with a geometric series. Like a ge- ometric series with |r| > 1, a series with ρ > 1 also diverges (Problem 30). However, if ρ = 1, the ratio test does not tell us anything; some series with ρ = 1 converge
14 Infinite Series, Power Series Chapter 1
and some diverge, so we must find another test (say one of the two preceding tests). To summarize the ratio test:
(6.3) If
ρ < 1 , the series converges; ρ = 1, use a different test; ρ > 1 , the series diverges.
Example 1. Test for convergence the series
n!
Using (6.2), we have
ρn =
∣∣^1
(n + 1)!
÷
n!
n! (n + 1)!
n(n − 1) · · · 3 · 2 · 1 (n + 1)(n)(n − 1) · · · 3 · 2 · 1
n + 1
ρ = lim n→∞ ρn = lim n→∞
n + 1
Since ρ < 1, the series converges.
Example 2. Test for convergence the harmonic series
n
We find
ρn =
n + 1
÷
n
∣ =^
n n + 1
ρ = lim n→∞
n n + 1
= lim n→∞
1 + (^) n^1
Here the test tells us nothing and we must use some different test. A word of warning from this example: Notice that ρn = n/(n + 1) is always less than 1. Be careful not to confuse this ratio with ρ and conclude incorrectly that this series converges. (It is actually divergent as we proved by the integral test.) Remember that ρ is not the same as the ratio ρn = |an+1/an|, but is the limit of this ratio as n → ∞.
PROBLEMS, SECTION 6
Use the ratio test to find whether the following series converge or diverge:
X^ ∞
n=
2 n n^2
X^ ∞
n=
3 n 22 n^
X^ ∞
n=
n! (2n)!
16 Infinite Series, Power Series Chapter 1
which we recognize (say by integral test) as a convergent series. Hence we use test (a) to try to show that the given series converges. We have:
lim n→∞
an bn
= lim n→∞
2 n^2 − 5 n + 1 4 n^3 − 7 n^2 + 2
÷
n^2
= lim n→∞
n^2
2 n^2 − 5 n + 1 4 n^3 − 7 n^2 + 2
= lim n→∞
2 − (^) n^5 + (^) n^12 4 − (^) n^7 + (^) n^23
Since this is a finite limit, the given series converges. (With practice, you won’t need to do all this algebra! You should be able to look at the original problem and see that, for large n, the terms are essentially 1/n^2 , so the series converges.)
Example 2. Test for convergence ∑^ ∞
n=
3 n^ − n^3 n^5 − 5 n^2
Here we must first decide which is the important term as n → ∞; is it 3n^ or n^3? We can find out by comparing their logarithms since ln N and N increase or decrease together. We have ln 3n^ = n ln 3, and ln n^3 = 3 ln n. Now ln n is much smaller than n, so for large n we have n ln 3 > 3 ln n, and 3n^ > n^3. (You might like to compute 100^3 = 10^6 , and 3^100 > 5 × 1047 .) The denominator of the given series is approximately n^5. Thus the comparison series is
n=2 3 n/n (^5). It is easy to prove this divergent by the ratio test. Now by test (b)
lim n→∞
3 n^ − n^3 n^5 − 5 n^2
÷
3 n n^5
= lim n→∞
1 − n
3 3 n 1 − (^) n^53
which is greater than zero, so the given series diverges.
PROBLEMS, SECTION 6
Use the special comparison test to find whether the following series converge or diverge.
X^ ∞
n=
(2n + 1)(3n − 5) √ n^2 − 73
X^ ∞
n=
n(n + 1) (n + 2)^2 (n + 3)
X^ ∞
n=
1 2 n^ − n^2
X^ ∞
n=
n^2 + 3n + 4 n^4 + 7n^3 + 6n − 3
X^ ∞
n=
(n − ln n)^2 5 n^4 − 3 n^2 + 1
X^ ∞
n=
√ n^3 + 5n − 1 n^2 − sin n^3
- Prove the special comparison test. Hint (part a): If an/bn → L and M > L, then an < M bn for large n. Compare
P∞ n=1 an^ with^
P∞ n=1 M bn.
Section 7 Alternating Series 17
7. ALTERNATING SERIES
So far we have been talking about series of positive terms (including series of abso- lute values). Now we want to consider one important case of a series whose terms have mixed signs. An alternating series is a series whose terms are alternately plus and minus; for example,
(7.1) 1 −
(−1)n+ n
is an alternating series. We ask two questions about an alternating series. Does it converge? Does it converge absolutely (that is, when we make all signs positive)? Let us consider the second question first. In this example the series of absolute values 1 +
n
is the harmonic series (6.1), which diverges. We say that the series (7.1) is not absolutely convergent. Next we must ask whether (7.1) converges as it stands. If it had turned out to be absolutely convergent, we would not have to ask this question since an absolutely convergent series is also convergent (Problem 9). However, a series which is not absolutely convergent may converge or it may diverge; we must test it further. For alternating series the test is very simple:
Test for alternating series. An alternating series converges if the absolute value of the terms decreases steadily to zero, that is, if |an+1| ≤ |an| and limn→∞ an = 0.
In our example
n + 1
n
, and lim n→∞
n
= 0, so (7.1) converges.
PROBLEMS, SECTION 7
Test the following series for convergence.
X^ ∞ n=
(−1)n √ n
X^ ∞ n=
(−2)n n^2
X^ ∞ n=
(−1)n n^2
X^ ∞
n=
(−3)n n!
X^ ∞
n=
(−1)n ln n
X^ ∞
n=
(−1)nn n + 5
X^ ∞
n=
(−1)nn 1 + n^2
X^ ∞
n=
(−1)n
√ 10 n n + 2
- Prove that an absolutely convergent series
P∞ n=1 an^ is convergent.^ Hint:^ Put^ bn^ = an + |an|. Then the bn are nonnegative; we have |bn| ≤ 2 |an| and an = bn − |an|.
- The following alternating series are divergent (but you are not asked to prove this). Show that an → 0. Why doesn’t the alternating series test prove (incorrectly) that these series converge?
(a) 2 − 1 2
2 3 − 1 4
2 5 − 1 6
2 7 − 1 8 · · ·
(b) 1 √ 2
− 1 2
1 √ 3
− 1 3
1 √ 4
− 1 4
1 √ 5
− 1 5 · · ·
Section 9 Useful Facts About Series 19
charges because there are an infinite number of them. At any stage the forces which would arise from the positive charges that are not yet in place, form a divergent series; similarly, the forces due to the unplaced negative charges form a divergent series of the opposite sign. We cannot then stop at some point and say that the rest of the series is negligible as we could in the bouncing ball problem in Section
- But if we specify the order in which the charges are to be placed, then the sum S of the series is determined (S is probably different from F in (8.1) unless the charges are placed alternately). Physically this means that the value of the force as the crews proceed comes closer and closer to S, and we can use the sum of the (properly arranged) infinite series as a good approximation to the force.
9. USEFUL FACTS ABOUT SERIES
We state the following facts for reference:
- The convergence or divergence of a series is not affected by multiplying every term of the series by the same nonzero constant. Neither is it affected by changing a finite number of terms (for example, omitting the first few terms).
- Two convergent series
n=1 an^ and^
n=1 bn^ may be added (or subtracted) term by term. (Adding “term by term” means that the nth term of the sum is an + bn.) The resulting series is convergent, and its sum is obtained by adding (subtracting) the sums of the two given series.
- The terms of an absolutely convergent series may be rearranged in any order without affecting either the convergence or the sum. This is not true of conditionally convergent series as we have seen in Section 8.
PROBLEMS, SECTION 9
Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don’t forget the preliminary test. Use the facts stated above when they apply.
X^ ∞
n=
n − 1 (n + 2)(n + 3)
X^ ∞
n=
n^2 − 1 n^2 + 1
X^ ∞
n=
1 nln 3
X^ ∞
n=
n^2 n^3 + 4
X^ ∞
n=
n n^3 − 4
X^ ∞
n=
(n!)^2 (2n)!
X^ ∞
n=
(2n)! 3 n(n!)^2
X^ ∞
n=
n^5 5 n^
X^ ∞
n=
nn n!
X^ ∞
n=
(−1)n^ n n − 1
X^ ∞
n=
2 n n^2 − 9
X^ ∞
n=
1 n^2 − n
X^ ∞
n=
n (n^2 + 4)^3 /^2
X^ ∞
n=
(−1)n n^2 − n
X^ ∞
n=
(−1)nn! 10 n
X^ ∞
n=
2 + (−1)n n^2 + 7
X^ ∞
n=
(n!)^3 (3n)!
X^ ∞
n=
(−1)n 2 ln^ n
20 Infinite Series, Power Series Chapter 1
1 22 − 1 32
1 23 − 1 33
1 24 − 1 34
1 2 +^1 22 − 1 3 − 1 32
1 4 +^1 42 − 1 5 − 1 52
X^ ∞
n=
an if an+1 = n 2 n + 3 an
- (a)
X^ ∞
n=
1 3 ln^ n^ (b)
X^ ∞
n=
1 2 ln^ n
(c) For what values of k is
X^ ∞
n=
1 kln^ n^
convergent?
10. POWER SERIES; INTERVAL OF CONVERGENCE
We have been discussing series whose terms were constants. Even more important and useful are series whose terms are functions of x. There are many such series, but in this chapter we shall consider series in which the nth term is a constant times xn^ or a constant times (x − a)n^ where a is a constant. These are called power series, because the terms are multiples of powers of x or of (x − a). In later chapters we shall consider Fourier series whose terms involve sines and cosines, and other series (Legendre, Bessel, etc.) in which the terms may be polynomials or other functions. By definition, a power series is of the form ∑^ ∞
n=
anxn^ = a 0 + a 1 x + a 2 x^2 + a 3 x^3 + · · · or
∑^ ∞
n=
an(x − a)n^ = a 0 + a 1 (x − a) + a 2 (x − a)^2 + a 3 (x − a)^3 + · · · ,
where the coefficients an are constants. Here are some examples:
1 −
x 2
x^2 4
x^3 8
(−x)n 2 n^
(10.2a) + · · · ,
x −
x^2 2
x^3 3
x^4 4
(−1)n+1xn n (10.2b) + · · · ,
x − x^3 3!
x^5 5!
x^7 7!
(−1)n+1x^2 n−^1 (2n − 1)!
(10.2c) + · · · ,
(x + 2) √ 2
(x + 2)^2 √ 3
(x + 2)n √ n + 1
(10.2d) + · · ·.
Whether a power series converges or not depends on the value of x we are considering. We often use the ratio test to find the values of x for which a series converges. We illustrate this by testing each of the four series (10.2). Recall that in the ratio test we divide term n + 1 by term n and take the absolute value of this ratio to get ρn, and then take the limit of ρn as n → ∞ to get ρ.
Example 1. For (10.2a), we have
ρn =
(−x)n+ 2 n+^
÷
(−x)n 2 n
x 2
ρ =
x 2
