infinite series

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Chapter 1
INFINITE SERIES
Chap1
Thison-linechaptercontainsthematerialoninfiniteseries,extractedfromtheprinted
versionoftheSeventhEditionandpresentedinmuchthesameorganizationinwhich
itappearedintheSixthEdition. Itiscollectedherefortheconvenienceofinstructors
whowishtouseitasintroductorymaterialinplaceofthatintheprintedbook. Ithas
been lightly edited to remove detailed discussions involving complex variable theory
that would not be appropriate until later in a course of instruction. For Additional
Readings, see the printed text.
1.1 INTRODUCTION TO INFINITE SERIES
Sec1.1
Perhapsthemostwidelyusedtechniqueinthephysicist’stoolboxistheuseofinfinite
series (i.e. sums consisting formally of an infinite number of terms) to represent
functions, to bring them to forms facilitating further analysis, or even as a prelude
to numerical evaluation. The acquisition of skill in creating and manipulating series
expansions is therefore an absolutely essential part of the training of one who seeks
competenceinthemathematicalmethodsofphysics,anditisthereforethefirsttopic
inthistext. Animportantpartofthisskillsetistheabilitytorecognizethefunctions
represented by commonly encountered expansions, and it is also of importance to
understand issues related to the convergence of infinite series.
FUNDAMENTAL CONCEPTS
The usual way of assigning a meaning to the sum of an infinite number of terms is
by introducing the notion of partial sums. If we have an infinite sequence of terms
u , u , u , u , u ,..., we define the i-th partial sum as
1 2 3 4 5
(cid:88)i
s = u . (1.1) eq1.1
i n
n=1
This is a finite summation and offers no difficulties. If the partial sums s converge
i
to a finite limit as i→∞,
lim s =S , (1.2) eq1.2
i
i→∞
(cid:80)
the infinite series ∞ u is said to be convergent and to have the value S. Note
n=1 n
thatwedefinetheinfiniteseriesasequaltoS andthatanecessaryconditionforcon-
1
2 CHAPTER 1. INFINITE SERIES
vergence to a limit is that lim u =0. This condition, however, is not sufficient
n→∞ n
to guarantee convergence.
SometimesitisconvenienttoapplytheconditioninEq.(1.2)inaformcalledthe
Cauchy criterion, namely that for each ε>0 there is a fixed number N such that
|s −s | < ε for all i and j greater than N. This means that the partial sums must
j i
cluster together as we move far out in the sequence.
Someseries diverge, meaningthatthe sequenceofpartialsumsapproaches±∞;
others may have partial sums that oscillate between two values, as for example
(cid:88)∞
u =1−1+1−1+1−···−(−1)n+··· .
n
n=1
This series does not converge to a limit, and can be called oscillatory. Often the
term divergent is extended to include oscillatory series as well. It is important to be
able to determine whether, or under what conditions, a series we would like to use is
convergent.
Example 1.1.1. The Geometric Series
The geometric series, starting with u = 1 and with a ratio of successive terms
0
r =u /u , has the form
n+1 n
1+r+r2+r3+···+rn−1+··· .
Its n-th partial sum s (that of the first n terms) is1
n
1−rn
s = . (1.3) eq1.3
n 1−r
Restricting attention to |r|<1, so that for large n, rn approaches zero, s possesses
n
the limit
1
lim s = , (1.4) eq1.4
n→∞ n 1−r
showing that for |r| < 1, the geometric series converges. It clearly diverges (or is
oscillatory) for |r|≥1, as the individual terms do not then approach zero at large n.
(cid:165)
Example 1.1.2. The Harmonic Series
As a second and more involved example, we consider the harmonic series
(cid:88)∞ 1 1 1 1 1
=1+ + + +···+ +··· . (1.5) eq1.5
n 2 3 4 n
n=1
The terms approach zero for large n, i.e. lim 1/n = 0, but this is not sufficient
n→∞
to guarantee convergence. If we group the terms (without changing their order) as
(cid:181) (cid:182) (cid:181) (cid:182) (cid:181) (cid:182)
1 1 1 1 1 1 1 1 1
1+ + + + + + + + +···+ +··· ,
2 3 4 5 6 7 8 9 16
(cid:80)
1Multiplyanddividesn= nm−=10rm by1−r.
1.1. INFINITE SERIES 3
each pair of parentheses encloses p terms of the form
1 1 1 p 1
+ +···+ > = .
p+1 p+2 p+p 2p 2
Forming partial sums by adding the parenthetical groups one by one, we obtain
3 4 5 n+1
s =1, s = , s > , s > ,..., s > ,
1 2 2 3 2 4 2 n 2
and we are forced to the conclusion that the harmonic series diverges.
Although the harmonic series diverges, its partial sums have relevance among
(cid:80)
other places in number theory, where H = n m−1 are sometimes referred to as
n m=1
harmonic numbers.
(cid:165)
Wenowturntoamoredetailedstudyoftheconvergenceanddivergenceofseries,
considering here series of positive terms. Series with terms of both signs are treated
later.
COMPARISON TEST
If term by term a series of terms u satisfies 0 ≤ u ≤ a , where the a form a
(cid:80)n n n n
convergent series, then the series u is also convergent. Letting s and s be
n n (cid:80) i j
partial sums of the u series, with j > i, the difference s −s is j u , and
j i n=i+1 n
this is smaller than the corresponding quantity for the a series, thereby proving
convergence. A similar argument shows that if term by term a series of terms v
(cid:80) n
satisfies 0 ≤ b ≤ v , where the b form a divergent series, then v is also
n n n n n
divergent.
For the convergent series a we already have the geometric series, whereas the
n
harmonic series will serve as the divergent comparison series b . As other series are
n
identifiedaseitherconvergentordivergent,theymayalsobeusedastheknownseries
for comparison tests.
Example 1.1.3. A Divergent Series
(cid:80)
Test ∞ n−p, p = 0.999, for convergence. Since n−0.999 > n−1 and b = n−1
n=1 (cid:80) n
forms the divergent harmonic series, the comparison test shows that n−0.999 is
(cid:80) n
divergent. Generalizing, n−p is seen to be divergent for all p≤1.
n
(cid:165)
CAUCHY ROOT TEST
(cid:80)
If (a )1/n ≤r <1 for all sufficiently large n, with r independent of n, then a is
n (cid:80) n n
convergent. If (a )1/n ≥1 for all sufficiently large n, then a is divergent.
n n n
The language of this test emphasizes an important point: the convergence or
divergence of a series depends entirely upon what happens for large n. Relative to
convergence, it is the behavior in the large-n limit that matters.
The first part of this test is verified easily by raising (a )1/n to the nth power.
n
We get
a ≤rn <1.
n
4 CHAPTER 1. INFINITE SERIES
(cid:80)
Since rn is just the nth term in a convergent geometric series, a is convergent
n n
by the comparison test. Conversely, if (a )1/n ≥1, then a ≥1 and the series must
n n
diverge. This root test is particularly useful in establishing the properties of power
series (Section 1.2).
D’ALEMBERT (OR CAUCHY) RATIO TEST
(cid:80)
If a /a ≤r <1 for all sufficiently large n and r is independent of n, then a
n+1 n (cid:80) n n
is convergent. If a /a ≥1 for all sufficiently large n, then a is divergent.
n+1 n n n
Thistestisestablishedbydirectcomparisonwiththegeometricseries(1+r+r2+
···). In the second part, a ≥ a and divergence should be reasonably obvious.
n+1 n
Although not quite as sensitive as the Cauchy root test, this D’Alembert ratio test
is one of the easiest to apply and is widely used. An alternate statement of the ratio
test is in the form of a limit: If

<1, convergence,
a
lim n+1 >1, divergence, (1.6) eq1.6
n→∞ an 
=1, indeterminate.
Because of this final indeterminate possibility, the ratio test is likely to fail at crucial
points, and more delicate, sensitive tests then become necessary. The alert reader
may wonder how this indeterminacy arose. Actually it was concealed in the first
statement, a /a ≤ r < 1. We might encounter a /a < 1 for all finite n but
n+1 n n+1 n
be unable to choose an r <1 and independent of n such that a /a ≤r for all
n+1 n
sufficiently large n. An example is provided by the harmonic series, for which
a n
n+1 = <1.
a n+1
n
Since
a
lim n+1 =1,
n→∞ an
no fixed ratio r <1 exists and the test fails.
Example 1.1.4. D’Alembert Ratio Test
(cid:80)
Test n/2n for convergence. Applying the ratio test,
n
a (n+1)/2n+1 1 n+1
n+1 = = .
a n/2n 2 n
n
Since
a 3
n+1 ≤ for n≥2,
a 4
n
we have convergence.
(cid:165)
CAUCHY (OR MACLAURIN) INTEGRAL TEST
Thisisanothersortofcomparisontest,inwhichwecompareaserieswithanintegral.
Geometrically, we compare the area of a series of unit-width rectangles with the area
under a curve.
1.1. INFINITE SERIES 5
Figure 1.1: (a) Comparison of integral and sum-blocks leading. (b) Comparison of
integral and sum-blocks lagging.
Fig1.1
Letf(x)beacontinuous,monotonic decreasing functioninwhichf(n)=a .
(cid:80) (cid:82) n
∞
Then a converges if f(x)dx is finite and diverges if the integral is infinite.
n n 1
The ith partial sum is
(cid:88)i (cid:88)i
s = a = f(n).
i n
n=1 n=1
But, because f(x) is monotonic decreasing, see Fig. 1.1(a),
(cid:90)
i+1
s ≥ f(x)dx.
i
1
On the other hand, as shown in Fig. 1.1(b),
(cid:90)
i
s −a ≤ f(x)dx.
i 1
1
Taking the limit as i→∞, we have
(cid:90) (cid:90)
∞ (cid:88)∞ ∞
f(x)dx≤ a ≤ f(x)dx+a . (1.7) eq1.7
n 1
1 n=1 1
Hencetheinfiniteseriesconvergesordivergesasthecorrespondingintegralconverges
or diverges.
This integral test is particularly useful in setting upper and lower bounds on the
remainder of a series after some number of initial terms have been summed. That is,
(cid:88)∞ (cid:88)N (cid:88)∞
a = a + a , (1.8) eq1.8
n n n
n=1 n=1 n=N+1
and
(cid:90) (cid:90)
∞ (cid:88)∞ ∞
f(x)dx≤ a ≤ f(x)dx+a . (1.9) eq1.9
n N+1
N+1 n=N+1 N+1
6 CHAPTER 1. INFINITE SERIES
To free the integral test from the quite restrictive requirement that the interpo-
lating function f(x) be positive and monotonic, we shall show that for any function
f(x) with a continuous derivative, the infinite series is exactly represented as a sum
of two integrals:
(cid:90) (cid:90)
(cid:88)N2 N2 N2
f(n)= f(x)dx+ (x−[x])f(cid:48)(x)dx. (1.10) eq1.10
n=N1+1 N1 N1
Here [x] is the integral part of x, i.e. the largest integer ≤ x, so x−[x] varies saw-
toothlike between 0 and 1. Equation ((1.10) is useful because if both integrals in
Eq. (1.10) converge, the infinite series also converges, while if one integral converges
and the other does not, the infinite series diverges. If both integrals diverge, the test
fails unless it can be shown whether the divergences of the integrals cancel against
each other.
We need now to establish Eq. (1.10). We manipulate the contributions to the
second integral as follows:
(1) Using integration by parts, we observe that
(cid:90) (cid:90)
N2 N2
xf(cid:48)(x)dx=N f(N )−N f(N )− f(x)dx.
2 2 1 1
N1 N1
(2) We evaluate
(cid:90) N2 N(cid:88)2−1 (cid:90) n+1 N(cid:88)2−1 (cid:104) (cid:105)
[x]f(cid:48)(x)dx= n f(cid:48)(x)dx= n f(n+1)−f(n)
N1 n=N1 n n=N1
(cid:88)N2
=− f(n)−N f(N )+N f(N ).
1 1 2 2
n=N1+1
Subtracting the second of these equations from the first, we arrive at Eq. (1.10).
An alternative to Eq. (1.10) in which the second integral has its sawtooth shifted
to be symmetrical about zero (and therefore perhaps smaller) can be derived by
methods similar to those used above. The resulting formula is
(cid:90) (cid:90)
(cid:88)N2 N2 N2
f(n)= f(x)dx+ (x−[x]− 1)f(cid:48)(x)dx
2
n=N1+1 N1 N1 (1.11) eq1.11
(cid:104) (cid:105)
+ 1 f(N )−f(N ) .
2 2 1
Because they do not use a monotonicity requirement, Eqs. (1.10) and (1.11) can
be applied to alternating series, and even those with irregular sign sequences.
Example 1.1.5. Riemann Zeta Function
The Riemann zeta function is defined by
(cid:88)∞
ζ(p)= n−p , (1.12) eq1.12
n=1
1.1. INFINITE SERIES 7
providing the series converges. We may take f(x)=x−p, and then
(cid:90) (cid:175)
∞ x−p+1 (cid:175)∞
x−p dx= (cid:175) , p(cid:54)=1,
(cid:175)
−p+1
1 x=1
(cid:175)
(cid:175)∞
=lnx(cid:175) , p=1.
x=1
The integral and therefore the series are divergent for p ≤ 1, and convergent for
p > 1. Hence Eq. (1.12) should carry the condition p > 1. This, incidentally, is an
independent proof that the harmonic series (p = 1) diverges logarithmically. The
(cid:80)
sum of the first million terms 1,000,000n−1 is only 14.392 726··· .
n=1
(cid:165)
While the harmonic series diverges, the combination
(cid:195) (cid:33)
(cid:88)n
γ = lim m−1−lnn (1.13) eq1.12a
n→∞
m=1
does converge, approaching a limit known as the Euler-Mascheroni constant.
Example 1.1.6. A Slowly Diverging Series
Exam1.1.6
Consider now the series
(cid:88)∞ 1
S = .
nlnn
n=2
We form the integral
(cid:90) ∞ 1 (cid:90) ∞ dlnx (cid:175)(cid:175)∞
dx= =lnlnx(cid:175) ,
2 xlnx x=2 lnx x=2
whichdiverges,indicatingthatS isdivergent. Noticethatthelowerlimitoftheinte-
gralisinfactunimportantsolongasitdoesnotintroduceanyspurioussingularities,
as it is the large-x behavior that determines the convergence. Because nlnn > n,
the divergence is slower than that of the harmonic series. But because lnn increases
more slowly than nε, where ε can have an arbitrarily small positive value, we have
(cid:80)
divergence even though the series n−(1+ε) converges.
n
(cid:165)
MORE SENSITIVE TESTS
Several tests more sensitive than those already examined are consequences of a the-
orem by Kummer. Kummer’s theorem, which deals with two series of finite positive
terms: u and a , states:
n n
(cid:80)
1. The series u converges if
n n
(cid:179) (cid:180)
u
lim a n −a ≥C >0, (1.14) eq1.13
n→∞ nun+1 n+1
whereC isaconstant. Thisstatementisequivalenttoasimplecomparisontest
(cid:80)
if the series a−1 converges, and imparts new information only if that sum
n n (cid:80)
diverges. The more weakly a−1 diverges, the more powerful the Kummer
n n
test will be.
8 CHAPTER 1. INFINITE SERIES
(cid:80)
2. If a−1 diverges and
n n
(cid:179) (cid:180)
u
lim a n −a ≤0, (1.15) eq1.14
n→∞ nun+1 n+1
(cid:80)
then u diverges.
n n
The proof of this powerful test is remarkably simple. Part 2 follows immediately
from the comparison test. To prove Part 1, write cases of Eq. (1.14) for n = N +1
through any larger n, in the following form:
u ≤(a u −a u )/C ,
N+1 N N N+1 N+1
u ≤(a u −a u )/C ,
N+2 N+1 N+1 N+2 N+2
...≤........................,
u ≤(a u −a u )/C .
n n−1 n−1 n n
Adding, we get
(cid:88)n a u a u
u ≤ N N − n n (1.16)
i C C
i=N+1
a u
< N N . (1.17) eq1.15
C
(cid:80)
This shows that the tail of the series u is bounded, and that series is therefore
n n
proved convergent when Eq. (1.14) is satisfied for all sufficiently large n.
Gauss’s Test is an application of Kummer’s theorem to series u > 0 when
n
the ratios of successive u approach unity and the tests previously discussed yield
n
indeterminate results. If for large n
u h B(n)
n =1+ + , (1.18) eq1.16
u n n2
n+1
(cid:80)
whereB(n)isboundedfornsufficientlylarge,thentheGaussteststatesthat u
n n
converges for h>1 and diverges for h≤1: There is no indeterminate case here.
The Gauss test is extremely sensitive, and will work for all troublesome series
the physicist is likely to encounter. To confirm it using Kummer’s theorem, we
(cid:80)
take a = nlnn. The series a−1 is weakly divergent, as already established in
n n n
Example 1.1.6.
Taking the limit on the left side of Eq. (1.14), we have
(cid:183) (cid:181) (cid:182) (cid:184)
h B(n)
lim nlnn 1+ + −(n+1)ln(n+1)
n→∞ n n2
(cid:183) (cid:184)
B(n)lnn
= lim (n+1)lnn+(h−1)lnn+ −(n+1)ln(n+1)
n→∞ n
(cid:183) (cid:181) (cid:182) (cid:184)
n+1
= lim −(n+1)ln +(h−1)lnn . (1.19) eq1.17
n→∞ n
For h<1, both terms of Eq. (1.19) are negative, thereby signalling a divergent case
of Kummer’s theorem; for h > 1, the second term of Eq. (1.19) dominates the first
1.1. INFINITE SERIES 9
and is positive, indicating convergence. At h=1, the second term vanishes, and the
first is inherently negative, thereby indicating divergence.
Example 1.1.7. Legendre Series
Ex1.1.7
TheseriessolutionfortheLegendreequation(encounteredinChapter7hassuccessive
terms whose ratio under certain conditions is
a 2j(2j+1)−λ
2j+2 = .
a (2j+1)(2j+2)
2j
To place this in the form now being used, we define u =a and write
j 2j
u (2j+1)(2j+2)
j = .
u 2j(2j+1)−λ
j+1
Inthelimitoflargej,theconstantλbecomesnegligible(inthelanguageoftheGauss
test, it contributes to an extent B(j)/j2, where B(j) is bounded). We therefore have
u 2j+2 B(j) 1 B(j)
j → + =1+ + . (1.20) eq1.18
u 2j j2 j j2
j+1
The Gauss test tells us that this series is divergent.
(cid:165)
Exercises
(cid:80)∞
1.1.1. (a) Provethatif lim npun =A<∞, p>1, theseries unconverges.
n→∞
n=1
(b) Prove that if lim nun =A>0, the series diverges. (The test fails for
n→∞
A=0.)
Thesetwotests,knownaslimit tests,areoftenconvenientforestablishingthe
convergence of a series. They may be treated as comparison tests, comparing
with (cid:88)
n−q, 1≤q <p.
n
b
1.1.2. If lim n =K, a constant with 0<K <∞, show that Σ b converges
n→∞an n n
or diverges with Σa .
n
b
Hint. If Σa converges, rescale b to b(cid:48) = n . If Σ a diverges, rescale to
n n n 2K n n
2b
b(cid:48)(cid:48) = n.
n K
(cid:88)∞ 1
1.1.3. (a) Show that the series converges.
n(lnn)2
n=2
10 CHAPTER 1. INFINITE SERIES
(cid:80)
(b) By direct addition 100,000[n(lnn)2]−1 =2.02288. Use Eq. (1.9) to make a
2
five-significant-figure estimate of the sum of this series.
1.1.4. Gauss’s test is often given in the form of a test of the ratio
u n2+a n+a
n = 1 0.
u n2+b n+b
n+1 1 0
For what values of the parameters a and b is there convergence? divergence?
1 1
ANS. Convergent for a −b >1,
1 1
divergent for a −b ≤1.
1 1
1.1.5. Test for convergence
(cid:88)∞ (cid:88)∞
(a) (lnn)−1 (d) [n(n+1)]−1/2
n=2 n=1
(cid:88)∞ n! (cid:88)∞ 1
(b) (e) .
10n 2n+1
n=1 n=0
(cid:88)∞ 1
(c)
2n(2n+1)
n=1
1.1.6. Test for convergence
(cid:181) (cid:182)
(cid:88)∞ 1 (cid:88)∞ 1
(a) (d) ln 1+
n(n+1) n
n=1 n=1
(cid:88)∞ 1 (cid:88)∞ 1
(b) (e) .
nlnn n·n1/n
n=2 n=1
(cid:88)∞ 1
(c)
n2n
n=1
(cid:88)∞ 1
1.1.7. For what values of p and q will converge?
np(lnn)q
n=2
(cid:40) (cid:40)
p>1, all q, p<1, all q,
ANS. Convergent for divergent for
p=1, q >1, p=1, q ≤1.
(cid:80)
1.1.8. Given 1,000n−1 = 7.485 470... set upper and lower bounds on the Euler-
n=1
Mascheroni constant.
ANS. 0.5767<γ <0.5778.
1.1.9. (From Olbers’ paradox.) Assume a static universe in which the stars are
uniformly distributed. Divide all space into shells of constant thickness; the
stars in any one shell by themselves subtend a solid angle of ω . Allowing for
0
the blocking out of distant stars by nearer stars, show that the total
net solid angle subtended by all stars, shells extending to infinity, is exactly
4π. [Therefore the night sky should be ablaze with light. For more details, see
E. Harrison, Darkness at Night: A Riddle of the Universe. Cambridge, MA:
Harvard University Press (1987).]