4. CONDITIONALLY CONVERGENT SERIES 23

4. Conditionally convergent series Here basic tests capable of detecting conditional convergence are studied.

4.1. Dirichlet’s test. Theorem 4.1. (Dirichlet’s test) n Let a complex sequence {An}, An = k=1 ak, be bounded, and the real sequence {bn} is decreasing monotonically, b1 ≥ b2 ≥ b3 ≥··· , so that P limn→∞ bn = 0. Then the series anbn converges. A proof is based on the identity:P m m m m−1

anbn = (An − An−1)bn = Anbn − Anbn+1 n=k n=k n=k n=k−1 X X X X m−1

= An(bn − bn+1)+ Akbk − Am−1bm n=k X Since {An} is bounded, there is a number M such that

|An|≤ M for all n. Therefore it follows from the above identity and non-negativity and monotonicity of bn that m m−1

anbn ≤ An(bn − bn+1) + |Akbk| + |Am−1bm| n=k n=k X X m−1

≤ M (bn − bn+1)+ bk + bm n=k ! X ≤ 2Mbk

By the hypothesis, bk → 0 as k → ∞. Therefore the right side of the above inequality can be made arbitrary small for all sufficiently large k and m ≥ k. By the Cauchy criterion the series in question converges. This completes the proof.

Example. By the root test, the series ∞ zn n n=1 X converges absolutely in the disk |z| < 1 in the complex plane. Let us investigate the convergence on the boundary of the disk. Put z = eiθ 24 1. THE THEORY OF CONVERGENCE

ikθ so that ak = e and bn = 1/n → 0 monotonically as n →∞. Then 1 − einθ A = eiθ + e2iθ + ··· + einθ = eiθ n 1 − eiθ 1 − einθ 1 − cos(nθ) 1 1 |A | = = ≤ ≤ n 1 − eiθ 1 − cos θ 1 − cos(θ) | sin(θ/2)| r s

Thus, the sequence {An} is bounded for all 0 <θ< 2π, and the series in question converges at all points of the closed disk |z| ≤ 1 except z = 1. The convergence on the boundary is not absolute.

4.2. Abel’s and Leibnitz tests. Dirichlet’s test provides a general test that allows one to detect conditional convergence. There are two con- sequences known as Abel’s and Leibnitz tests. They are also used to detect conditional convergence. Corollary 4.1. (Leibnitz criterion for alternating series) Suppose that |cn| is decreasing monotonically, |cn+1|≤|cn|, so that cn → 0 as n → ∞, and c2m−1 ≥ 0, c2m ≤ 0. Then the series cn converges. P n In Dirichlet’s test, put an = (−1) and bn = |cn|. Then |An| ≤ 1 and bn → 0 monotonically, Leibnitz criterion follows.

Alternating harmonic series. By Leibnitz criterion, the alternating har- monic series ∞ (−1)n+1 n n=1 X converges, but not absolutely. Theorem 4.2. (Abel’s test) Suppose a complex series an converges and a real sequence {cn} is monotonic and bounded. Then the series anbn converges. P n A proof of Abel’s test is also based onP Dirichlet’s test. A bounded monotonic sequence (either increasing or decreasing) converges. Sup- pose cn ≤ cn+1 for all n. Then there exists c such that cn → c as n → ∞ monotonically. Therefore the sequence bn = c − cn ≥ 0 is non-negative and converges to 0. The convergence of an means that the sequence of partial sums An = a1 + a2 + ··· + an converges. But any convergent sequence is bounded. Therefore by DirichletP ’s test, the series n an(c − cn) converges. From the convergence of an and P P 4. CONDITIONALLY CONVERGENT SERIES 25

n an(c − cn) and the limit laws it follows that the series n ancn converges and P P ancn = c an − an(c − cn) n n X X X Similarly, if cn ≥ cn+1 for all n. Then put bn = cn − c, where c is the limit of a bounded monotonically decreasing sequence cn, in Dirichlet’s test, and the convergence of ancn follows from the limit laws.

4.3. Rearrangements. ConsiderP a one-to-one map of the set of all pos- itive integer onto itself so that {1, 2, 3,...} is mapped to {k1, k2, k3,...}. The series 0 0 0 0 akn = ak1 + ak2 + ak3 + ··· = a1 + a2 + a3 + ··· = an n n X X is called a rearrangement of the series n an = a1 + a2 + a3 + ··· . Theorem 4.3. (Rearrangement andP absolute convergence) A rearrangement of any absolutely convergent series converges abso- lutely to the same sum: 0 an = an . n n X X The convergence of a rearrangement of an absolutely convergent series follows from the Cauchy criterion. Indeed, given ε > 0, there is an integer N such that m

|an| <ε, ∀m ≥ k ≥ N n=k 0 X Let Sn and Sn be partial sums of the series and its rearrangement, respectively. For any rearrangement there is an integer M ≥ N such that the first M terms of the rearrangement contains the terms a1, 0 a2,..., aN . Then the latter terms are canceled in the difference Sn − Sn 0 if n ≥ M ≥ N so that |Sn − Sn| does not exceed the sum of |aj| where j takes integer values no less than N. By the above inequality 0 0 |Sn − Sn| <ε for all n ≥ M. Therefore Sn converges to the same limit as Sn. In contrast, a rearrangement of a conditionally convergent series can converge to a different sum and even fail to converge. For example, consider the alternating harmonic series ∞ (−1)n+1 1 1 1 1 1 = 1 − + − + − + ··· = S n 2 3 4 5 6 n=1 X 26 1. THE THEORY OF CONVERGENCE

and let Sn be the sequence of its partial sums so that Sn → S as n →∞. Evidently, 1 1 5 S

1 1 1 0 0 0 + − > 0 ⇒ S3 = S

0 0 5 lim sup Sn >S3 = >S n→∞ 6 0 and, hence, the sequence Sn cannot converge to S. Theorem 4.4. (Properties of conditionally convergent series) + Let an be a real series that converges but not absolutely. If Sp is the − sum of the first p positive terms of the series and Sq is the sum of its firstPq negative terms, then + − lim Sp = ∞ , lim Sq = −∞ . p→∞ q→∞ For example, ∞ (−1)n+1 1 1 1 1 1 = 1 − + − + − + ··· n 2 3 4 5 6 n=1 X ∞ + 1 1 1 lim Sp = lim 1+ + ··· + = = ∞ p→∞ p→∞ 3 2p − 1 2p − 1   p=1 X∞ − 1 1 1 1 1 lim Sq = lim − − −···− = − = −∞ q→∞ q→∞ 2 4 2q 2 q q=1   X In general, the stated property follows from the following observa- tion. Consider two series n bn and n cn where 1 1 b = (a +P|a |) ,P c = (a −|a |) n 2 n n n 2 n n 4. CONDITIONALLY CONVERGENT SERIES 27

Clearly, bn = an if an > 0 and bn = 0 otherwise. Similarly, cn = an if an < 0 and cn = 0 otherwise. Let An, Bn, Cn, and Sn be the partial sums of an, bn, cn, and |an|, respectively. Then

P P P1 P 1 B = (A + S ) , C = (A − S ) n 2 n n n 2 n n

Since an converges but not absolutely, An converges to some number A but lim Sn = ∞. Therefore P

lim Bn = ∞ , lim Cn = −∞ n→∞ n→∞

+ − Evidently, Bn = Sp for any n and some p ≤ n, and Cn = Sq for any ± n and some q = n − p ≤ n (if S0 = 0 by definition). It is concluded that a real conditionally convergent series necessarily contains infinitely many negative terms and infinitely many positive terms (otherwise the corresponding partial sums Bn and Cn were finite). Therefore the limit n → ∞ implies that p → ∞ and q → ∞ and the conclusion of the theorem follows.

Remark. As the series made of positive and negative terms of a condi- tionally convergent series diverge to infinity, the sum of the series looks like an indeterminate form “∞−∞” whose value depends on how the infinities in this form are “regularized”, or, more precisely, how negative and positive terms are ordered when computing the sequence of partial sums. This characteristic property of conditionally convergent series is quantified by the following theorem.

Theorem 4.5. (Riemann)

Let n an converges but not absolutely. For any two elements of the extended real number system −∞ ≤ α ≤ β ≤ ∞, there exists a re- P 0 0 arrangement n an with partial sums Sn such that P 0 0 lim inf Sn = α, lim sup Sn = β n→∞ n→∞

The Riemann theorem about rearrangements shows that a rearrange- ment of a conditionally convergent series may be constructed to con- verge to any desired number α = β, or diverge to ±∞, or to have a sequence of partial sums that does not converge and oscillates between any two numbers (α<β) (or has unbounded oscillations, α = −∞, β = ∞, or both). 28 1. THE THEORY OF CONVERGENCE

4.4. A rearrangement converging to any designated number. To illustrate the Riemann theorem3, let us construct a rearrangement of a condition- ally convergent series that converges to a given number β > 0 (the case 0 β ≤ 0 can be considered along a similar line of arguments). Let Sn denote the sequence of partial sums of the sought-after rearrangement. 0 + ∞ The rearrangement should be such that Sn → β as n →∞. Let {am}1 − ∞ and {ak }1 denote the subsequences of positive and negative terms in the sequence {an}. Clearly, these sequences can be reordered so that they converge to 0 monotonically. Note that an → 0 as n →∞ owing to the conditional convergence of an. So, without loss of generality, + the positive subsequence {am}⊂{an} is monotonically decreasing to P − zero, while the negative subsequence {ak }⊂{an} is monotonically increasing to zero. + Suppose that a1 <β. The idea is + • to add positive terms of {am} until their sum overshoots β; − • then to add negative terms from {ak } until the partial sum becomes less than β; • then to add positive terms to make the partial sum greater than β again • then to add negative terms again to make the partial sum less than β and so on. The result is a sequence of partial sums that oscillates about β. • Finally, one shows that the amplitude of the oscillations tends to zero with increasing the number of terms in partial sums. 0 + Let us demonstrate that the above idea does work. Put an = an for n = 1, 2,...,m1 where m1 is determined by the condition 0 + + + 0 + 0 Sm1−1 = a1 + a2 + ··· + am1−1 <β ≤ Sm1−1 + am1 = Sm1 + Since the series am = ∞ diverges, the sum of finitely many terms of {a+ } can exceed any preassigned positive number β. This fulfills the m P first step of the above procedure. The next k1 terms of the rearrange- ment are taken from the negative subsequence {ak} and the integer k1 is determined by the condition that the corresponding partial sum become less than β: 0 − am1+k = ak , k = 1, 2,...,k1 0 0 Sm1+k1 <β ≤ Sm1+k1 −1 − The integer k1 always exists as ak = −∞. This comprises the second step of the above construction. P 3A proof the Riemann theorem can be found in: W. Rudin, Principles of Math- ematical Analysis 4. CONDITIONALLY CONVERGENT SERIES 29

0 + Next m2 terms of the rearrangement {an} are taken from {am} again and m2 is determined by the condition that the corresponding partial sum exceeds β again: 0 + am1+k1+j = am1+j , j = 1, 2,...,m2 0 0 Sm1+k1 +m2−1 <β ≤ Sm1+k1+m2 + Note that that m>m1 am = ∞ for any choice of m1 and, hence, the + above condition can be met with finitely many terms am, m>m1. The P 0 − next k2 terms of {an} are taken from {ak }, where k2 is determined by the condition that the corresponding partial sum drops below β again: 0 − am1+k1+m2+j = ak1+j , j = 1, 2,...,k2 0 0 Sm1+k1+m2+k2 <β ≤ Sm1+k1+m2+k2−1

The existence of the integer k2 is guaranteed by the divergence of the − series k>k1 ak = −∞ for any k1. 0 + Thus, the rearrangement {an} consists of the first m1 terms of {am}, P − + then the first k1 terms of {ak }, then next m2 terms of {am}, then next − k2 terms of {ak } and so on: 0 {an} = + + − − + + − − {a1 ,...,am1,a1 ,...,ak1,am1+1,...,am1+m2 ,ak1+1,...,ak1+k2 ,...} . 0 The sequence of partial sums {Sn} of the constructed rearrangement 0 oscillates about the number β. To prove that the sequence {Sn} converges to β, it is sufficient to show that the amplitude of oscil- 0 lations tends to zero with increasing n. The sequence {Sn} is in- creasing if n lies in the intervals [1,m1],[m1 + k1 + 1,m1 + k1 + m2], [m1 + k1 + m2 + k2 + 1,m1 + k1 + m2 + k2 + m3], etc. The sequence is decreasing if the index takes values in the intervals [m1 + 1,m1 + k1], [m1 + k1 + m2 + 1,m1 + k1 + m2 + k2], etc. Therefore the numbers 0 0 0 Sm1 − β, Sm1+k1+m2 − β, etc., define how much {Sn} overshoots β, 0 0 whereas the numbers β − Sm1+k1 , β − Sm1+k1+m2+k2 , etc., define how 0 much {Sn} undershoots β upon each oscillation. It follows from the above conditions on the integers m1, m2,... that + 0 0 + −am1 <β − Sm1 ≤ 0 ⇒ 0 ≤ Sm1 − β

0 ≤ Sm1+k1+···+mj−1+kj−1+mj − β

0 + where j = 1, 2,... (enumerates oscillations of {Sn}). The sequence {am} converges to zero monotonically and so do the “overshots”. Similarly, it follows from the above definitions of k1, k2, etc., that − 0 <β − Sm1+k1 ≤ −ak1 − 0 <β − Sm1+k1+m2+k2 ≤ −ak1+k2 − so that all “undershots” are bounded by terms of {ak }:

0 <β − Sm1+k1+···+mj+kj ≤−ak1+···+kj − The sequence {ak } converges to zero monotonically and so do the “un- dershots”. This means that the amplitude of oscillations of the se- quence of partial sums of the constructed rearrangement tends to zero and, hence, 0 lim Sn = β n→∞ as desired. So, the sum of a conditionally convergent series depends on the summation order. It is not difficult to implement the above algorithm numerically for any conditionally convergent sequence. In particu- lar, one can find a rearrangement of the alternating harmonic series (−1)n+1/n that converges to any preassigned number.

PExercises.

1. Investigate the convergence of the Fourier series ∞ einθ , p> 0 np n=1 X 2. Show that the series ∞

an sin(nx) n=1 X converges for all real x if an → 0 monotonically as n →∞.

3. Show that the series ∞

an cos(nx) n=1 X converges for all real x =6 2mπ, where m is an integer, if an → 0 mono- tonically as n →∞. 4. CONDITIONALLY CONVERGENT SERIES 31

4. Show that the series ∞ n (−1) an cos(nx) n=1 X converges for all real x =6 (2m − 1)π, where m is an integer, if an → 0 monotonically as n →∞.

5. Show that the series ∞ n (−1) an sin(nx) n=1 X converges for all real x, if an → 0 monotonically as n →∞.