Sequences and Series of Numbers

Lecture 21

Sequences and Series

Sequences and Series of Numbers 139 P6. If lim_n→∞z_n =z, then lim_n→∞z_n =z and lim_n→∞|z_n|=

|lim_n→∞z_n|=|z|.

P7. Ifz_n=x_n+iy_n, n≥0 andz=x+iy,then lim_n→∞z_n =z if and only ifxn→xandyn →y.

P8. If limn→∞zn = 0 and|wn| ≤ |zn|, n≥0, then limn→∞wn = 0.

P9. If limn→∞zn = 0 and {wn} is a bounded sequence, then limn→∞wnzn= 0.

P10. From any bounded sequence, it is possible to extract a convergent subsequence.

P11. A sequence{zn} is called aCauchy sequenceif for any >0 there exists an integerN >0 such thatn, m≥N ⇒ |zn−zm|< .The sequence {zn}is convergent if and only if it is a Cauchy sequence.

P12. (Heine’s Criterion). Letf be a function defined in a neighborhood ofz.The functionf is continuous atzif and only if for any sequence{z_n} converging to zthe condition limn→∞f(zn) =f(z) holds.

A point z is called a limit point (or anaccumulation point) of the se- quence {zn} if for every neighborhood N(z) there exists a subsequence {zn_k} for which all terms belong to N(z). Alternatively, z is said to be a limit point of{zn} if every neighborhoodN(z) contains infinitely many terms of{zn}. From P10, in particular, it follows that every bounded real sequence{an} has at least one limit point. For example, for the sequence {(−1)ⁿ}, −1 and 1 are the limit points. An unbounded sequence may not have a limit point; e.g. {√

n}; however, an unbounded real sequence {an} has at least one limit point if we allow the values +∞ and −∞. This means that the real sequence {an} has +∞ as a limit point if it contains arbitrarily large positive terms, and−∞as a limit point if it con- tains negative terms of arbitrarily large absolute value. Clearly, for the sequence{1,1,2,1,2,3,1,2,3,4,· · ·},each natural number is a limit point.

The largest (smallest) limit point of the (bounded or unbounded) real se- quence {an} is called the limit superior or upper limit (limit inferior or lower limit) and is denoted as lim sup_n→∞an (lim infn→∞an). For exam- ple, the sequence {an} defined by un = (−1)ⁿ(3 + 1/n) is bounded with lower bound −4 and upper bound 7/2.For this sequence, the limit points are −3 and 3, and hence lim sup_n→∞an = 3 and lim infn→∞an = −3.

For the sequence {1,1,2,1,2,3,1,2,3,4,· · ·}, lim sup_n→∞an = ∞ and lim infn→∞an = 1. For a real bounded sequence {an}, it follows that limn→∞an = a if and only if lim infn→∞an = a = lim sup_n→∞an. For an unbounded real sequence{an}, we have limn→∞an =∞ if and only if lim infn→∞an=∞= lim sup_n→∞an,and limn→∞an=−∞if and only if lim infn→∞an=−∞= lim sup_n→∞an.Now let{an}and{bn}be bounded sequences of real numbers. For these sequences, it can easily be shown that:

(1). If a_n ≤ b_n, n ∈ N, then lim sup_n→∞a_n ≤ lim sup_n→∞b_n and lim inf_n→∞a_n≤lim inf_n→∞b_n.

(2). lim sup_n→∞(a_n+b_n)≤lim sup_n→∞a_n+ lim sup_n→∞b_n. (3). lim infn→∞(an+bn)≥lim infn→∞an+ lim infn→∞bn.

In (2) and (3), the inequality sign cannot be replaced by equality. For this it suffices to considera_n = (−1)ⁿ andb_n= (−1)ⁿ⁺¹.

A series is a formal expression of the form z₀+z₁+· · · or, equiva- lently, $_∞

j=0zj,where the termszj are complex numbers. Thenthpartial sum of the series, denoted by sn, is the sum of the first n+ 1 terms; i.e., sn =$n

j=0zj. If the sequence of partial sums {sn}^∞n=0 has a limit s, the series is said toconverge, orsum, to s,and we writes=$_∞

j=0zj.A series that does not converge is said to diverge. The series $_∞

j=0zj is said to be absolutely convergent provided that the series of magnitudes $_∞

j=0|zj| converges. For example, the series $_∞

j=0(−i)^j/(j + 1)² converges abso- lutely. A complex series that is convergent but not absolutely convergent is calledconditionally convergent. For example, the series$_∞

j=0(−1)^j/(j+ 1) converges conditionally.

Example 21.3.

Consider the geometric series$_∞

j=0ac^j,where|c|<1.

From Lemma 19.2, we have a

1−c −(a+ac+ac²+· · ·+acⁿ) = acⁿ⁺¹ 1−c.

Since|c|<1,the term|a||c|ⁿ⁺¹/|1−c|converges to 0 asn→ ∞.Thus, the geometric series converges to a/(1−c).

The following properties of complex series are similar to those of real series:

Q1. If$_∞

j=0zj converges, then limn→∞zn= 0.

Q2. If$_∞

j=0zj converges, then limm→∞$_∞

n=m+1zn= 0.

Q3. If$_∞

j=0z_jconverges, then there exists a real constantM such that

|z_j| ≤M for allj.

Q4. If$_∞

j=0zj and$_∞

j=0wj converge, andαand βare complex num- bers, then$_∞

j=0(αzj+βwj) =α$_∞

j=0zj+β$_∞

j=0wj. Q5. Ifz_j=x_j+iy_j ands=u+iv,then$_∞

j=0z_jconverges tosif and only if $_∞

j=0x_j=uand$_∞

j=0y_j=v.

Q6. If $_∞

j=0z_j converges absolutely, then it converges; however, the converse is not true.

Q7. If$_∞

j=0zj converges absolutely, then its every rearrangement is absolutely convergent and converges to the same limit.

Q8. The series $_∞

j=0zj converges if and only if for any given > 0 there exists anN >0 such that m > n≥N implies|s_m−s_n|=|z_n+1+

Sequences and Series of Numbers 141 z_n+2+· · ·+z_m|< .This is known asCauchy’s criterion.

Q9. (Comparison Test). If|z_j| ≤ a_j for all j and$_∞

j=0a_j converges, then$_∞

j=0z_j converges.

Q10. (Cauchy’s Root Test). If ρ = limj→∞|zj|^1/j either exists or is infinite, then the series$_∞

j=0zj converges absolutely ifρ <1 and diverges ifρ >1.Ifρ= 1 the test is inconclusive.

Q11. (d’Alembert’s Ratio Test). If limj→∞|c_j+1/cj|=ρ,then the series

$_∞

j=0zj converges absolutely if ρ <1 and diverges if ρ >1. If ρ= 1 the test is inconclusive.

Now let$_∞

j=0z_j and $_∞

j=0w_j be given series. TheCauchy product of these series is defined to be the new series$_∞

j=0t_j,where

tj = z₀wj+z₁wj−1+· · ·+zj−1w₁+zjw₀ = j k=0

zkwj−k = j k=0

zj−kwk. (21.2) The following example shows that the Cauchy product of two convergent series need not be convergent.

Example 21.4.

The series $_∞

j=0zj and $_∞

j=0wj, where zj = wj = (−1)^j(j+ 1)^−1/2, are convergent. The Cauchy product of these series is

$_∞

j=0tj, where

tj = (−1)^j j k=0

(k+ 1)^−1/2(j+ 1−k)^−1/2. Now since from the arithmetic-geometric mean inequality

(k+ 1)(j+ 1−k) ≤ k+ 1 +j+ 1−k

2 = j+ 2

2 , we find

(k+ 1)^−1/2(j+ 1−k)^−1/2 ≥ 2 j+ 2. Thus, it follows that

(−1)^jtj ≥ (j+ 1) 2

j+ 2 > 1 for all j >0.

Hence, t_j does not tend to zero as j → ∞. This shows that the Cauchy product of these series diverges.

Q12. If $_∞

j=0z_j and $_∞

j=0w_j converge to z and w, respectively, and their Cauchy product converges, then

∞ j=0

tj =

⎛

⎝^∞

j=0

zj

⎞

⎠

⎛

⎝^∞

j=0

wj

⎞

⎠ = zw. (21.3)

However, if $_∞

j=0z_j and$_∞

j=0w_j converge absolutely to zand w,respec- tively, then their Cauchy product also converges absolutely, and (21.3) holds.

Problems

21.1. Discuss the convergence of the following sequences:

(a).

3n+ 2(i)ⁿ n

, (b).

1 +i

√2 n

, (c).

3n+ 7ni 2n+ 5i

. 21.2. Discuss the convergence of the following series:

(a).

∞ j=0

i^j

(j+ 1)², (b).

∞ j=0

(1 + 3i)^j 5^j , (c).

∞ j=0

cosjπ

5 +isinjπ 5

, (d).

∞ j=0

2i^j 5 +ij². 21.3. Examine the series $_∞

j=0zj for its convergence or divergence, wherez_j are recursively defined as

z0= 1 +i, zj+1 = (2 + 3i)j 7 + 5ij²zj. 21.4. If 0≤r <1,show that

(a).

∞ j=0

r^jcosjθ= 1−rcosθ

1 +r²−2rcosθ, (b).

∞ j=0

r^jsinjθ= rsinθ 1 +r²−2rcosθ. 21.5. If lim_n→∞z_n=z,show that lim_n→∞(z₀+z₁+· · ·+z_n)/(n+ 1) = z.

21.6. Prove that if$_∞

j=0zj converges and|argzj| ≤θ < π/2, then it converges absolutely.

21.7. Suppose that$_∞

j=0z_jand$_∞

j=0z_j²are convergent and Rez_j ≥0.

Show that$_∞

j=0|z_j|² converges.

Sequences and Series of Numbers 143 21.8. A metric space (S, d) is said to be complete if every Cauchy sequence inS converges. Show that (C, d) with any metricdis complete.

21.9. The functionf :S→Cis said to be uniformly continuousonS if for every given >0 there exists a δ =δ()>0 such that|z−w| < δ implies|f(z)−f(w)|< for allz, w∈S. Show that:

(a). The functionf(z) = 1/zis continuous on S ={z: 0 <|z|<1},but not uniformly continuous.

(b). If S is compact and f(z) is continuous on S, then it is uniformly continuous.

Answers or Hints

21.1. (a). Converges to 3, (b). in polar form limn→∞e^{i nπ/4}oscillates, and hence diverges, (c). converges to (3 + 7i)/2.

21.2. (a). Absolutely convergent, (b). geometric series 5/(4−3i),(c). di- vergent, (d). absolutely convergent.

21.3. Absolutely convergent.

21.4. In Example 21.3, leta= 1, c=re^iθ.

21.5. Writingwn=zn−z,this is equivalent to showing that if limwn = 0, then lim(w0+w1+· · ·+wn)/(n+ 1) = 0.Given ε >0,there areN1≥1 andN2 such thatn≥N1implies|wn|< ε/2 and

n≥N₂⇒ ^|w0+w1+_n+1^{···+wN1−1|} <₂^ε. Then, by the triangle inequality,

n≥max{N1, N2} ⇒ ^|w0+w_n+1^1+···+wn| ≤^|w0+w1+_n+1^{···+wN1−1|} +^|wN1|_n+1^+···+|wn|< ^ε₂+ⁿ⁻_n+1^{N1+1 ε}₂ < ε.

21.6. Writing zj =xj+iyj = rje^iθj, since xj >0, |yj| =xj|tanθj| ≤ x_j tanθ, and$

x_j converges,$

|y_j|also converges. Since|z_j| ≤ |x_j|+|y_j| by the triangle inequality, then $

|zj|converges.

21.7. Writing z_j = x_j +iy_j, since $

x_j converges and x_j ≥ 0, {x_j} is bounded, say, xj ≤ M. Then 0 ≤ x²_j ≤ M xj, so $

x²_j also converges.

Sincez²_j =x²_j−y²_j+ 2ixy,$

(x²_j−y_j²) converges. Then so does$

y_j²since y_j²=x²_j−(x²_j−y²_j). Since|z_j|²=x²_j+y²_j, then$

|z_j|²also converges.

21.8. We will show that C with the Euclidean metric is complete. Let {z_n}be a Cauchy sequence inC,so that for >0 there is anN such that

|z_n−z_m|< forn, m≥N.Letz_n=x_n+iy_n.Since|x_n−x_m|<|z_n−z_m| and|y_n−y_m|<|z_n−z_m|for all n, m,we find that{x_n}and{y_n}are real Cauchy sequences. They both converge since IR is complete, say topand q,respectively. Then, it follows that{z_n} converges top+iq∈C.

21.9. (a). Let z = δ (0 < δ <1) and =δ/(1 +). Then, |z−w| < δ, but |f(z)−f(w)|=/δ > .(b). Suppose not. Then there exists az₀,and for eachnthere are pointsz_n andw_n in S such that|z_n−w_n|<1/nand

|f(z_n)−f(w_n)| ≥₀. Now, sinceS is compact, there exists a subsequence z_n1, z_n2,· · · and a point z_∗ ∈ S such that z_∗ = lim_k→∞z_nk. Since |z_∗− w_nk| ≤ |z_∗−z_nk|+ 1/n_k,it is clear thatz_∗= lim_k→∞w_nk.Thus, from the continuity off(z),we havef(z_∗) = lim_k→∞f(z_nk) = lim_k→∞f(w_nk).Now notice that in₀≤ |f(z_nk)−f(w_nk)| ≤ |f(z_nk)−f(z_∗)|+|f(z_∗)−f(w_nk)| the right-hand side tends to zero.

Lecture 22

Sequences and Series