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Numerical Sequences and Series
Written by Men-Gen Tsai email: b89902089@ntu.edu.tw
- Prove that the convergence of {sn} implies convergence of {|sn|}. Is the converse true? Solution: Since {sn} is convergent, for any > 0, there exists N such that |sn − s| < whenever n ≥ N. By Exercise 1.13 I know that ||sn| − |s|| ≤ |sn − s|. Thus, ||sn| − |s|| < , that is, {sn} is convergent. The converse is not true. Consider sn = (−1)n.
- Calculate limn→∞ (
n^2 + n − n). Solution: √ n^2 + n − n = √ n n^2 + n + n = √^1 1 /n + 1 + 1 → (^12)
as n → ∞.
- If sn =
2 and
sn+1 =
√ 2 + √sn (n = 1, 2 , 3 , ...),
prove that {sn} converges, and that sn < 2 for n = 1, 2 , 3 , .... Proof: First, I show that {sn} is strictly increasing. It is trivial that s 2 =
√ 2 + √s 1 =
√ 2 +
2 = s 1. Suppose sk > sk− 1 when
k < n. By the induction hypothesis,
sn =
√ 2 + √sn− 1
√ 2 + √sn− 2 = sn− 1
By the induction, {sn} is strictly increasing. Next, I show that {sn} is bounded by 2. Similarly, I apply the induction again. Hence {sn} is strictly increasing and bounded, that is, {sn} converges.
- Prove that the convergence of ∑^ an implies the convergence of
∑ √an n if an ≥ 0. Proof: By Cauchy’s inequality, ∑^ k n=
an
∑k n=
n^2 ≥
∑^ k n=
an
√a n n
for all n ∈ N. Also, both ∑^ an and ∑^ n^12 are convergent; thus ∑kn=1 an
√an n is bounded. Besides,
√an n ≥^ 0 for all^ n. Hence^
∑ √an n is convergent.
- Find the radius of convergence of each of the following power series:
(a)
∑ n^3 zn, (b)
∑ 2 n n! z
n, (c) ∑^2 n n^2 z
n, (d) ∑^ n^3 3 n^ z
n.
all n > m ≥ N 2. Take N = max(N 1 , N 2 ). Thus
> (^) 1 +am a m
(^) 1 + 1 am + ... + (^) 1 + 1an
= am^ +^ ... 2 +^ an
for all n > m ≥ N. Thus
am + ... + an < 2
for all n > m ≥ N. It is a contradiction. Hence ∑^ 1+anan diverges. Proof of (b):
aN + sN +1^ +^ ...^ +^
aN +k sN +k^ ≥^
aN + sN +k^ +^ ...^ +^
aN +k sN +k = aN^ +1^ +^ ...^ +^ aN^ +k sN +k = sN^ + sk^ −^ sN N +k = 1 − (^) ssN N +k
If ∑^ a snn converges, for any > 0 there exists N such that
am sm^ +^ ...^ +^
an sn^ <
for all m, n whenever n > m ≥ N. Fix m = N and let n = N + k. Thus
> a sm m
for all k ∈ N. But sN +k → ∞ as k → ∞ since ∑^ an diverges and an > 0. Take = 1/2 and we obtain a contradiction. Hence ∑^ a snn diverges.
Proof of (c):
sn− 1 ≤ sn ⇔ (^) s^12 n
≤ (^) s^1 nsn− 1 ⇔ a sn 2 n
≤ (^) san nsn− 1
= sn s^ −^ sn−^1 nsn− 1 ⇔ a sn 2 n
≤ (^) s^1 n− 1
− (^) s^1 n for all n.
Hence ∑^ k n=
an s^2 n^ ≤
∑^ k n=
( (^) s^1 n− 1
− (^) s^1 n
= (^) s^1 1
− (^) s^1 n
Note that (^) s^1 n → 0 as n → ∞ since ∑^ an diverges. Hence ∑^ a s (^2) nn converges.
Proof of (d): ∑^ 1+anann may converge or diverge, and ∑^ 1+ann (^2) an con- verges. To see this, we put an = 1/n. (^) 1+anann = (^21) n , that is, ∑^ 1+anann = 2 ∑^1 /n diverges. Besides, if we put
an = (^) n(log^1 n)p
where p > 1 and n ≥ 2, then an 1 + nan^ =^
n(log n)^2 p((log n)p^ + 1) < 1 2 n(log n)^3 p for large enough n. By Theorem 3.25 and Theorem 3.29, ∑^ 1+anann converges. Next, ∑ (^) an 1 + n^2 an^ =^
1 /an + n^2
for all n > N. But rn → 0 as n → ∞; thus a rmm + ... + a rnn → 1 as n → ∞. If we take = 1/2, we will get a contradiction.
Proof of (b): Note that
rn+1 < rn ⇔ √rn+1 <
rn ⇔
rn + √rn+1 < 2
rn ⇔
√r n +^ √r √ n+ rn
rn − √rn+1)
√r n +^ √r √ n+ rn^ <^ 2(
rn − √rn+1)
⇔ rn^ − √^ rrn+ n
rn − √rn+1)
⇔ √arn n
< 2(√rn − √rn+1)
since an > 0 for all n.
Hence,
∑^ k n=
√^ an rn^ <
∑^ k n=
2(√rn − √rn+1) = 2(√r 1 − √rk+1)
Note that rn → 0 as n → ∞. Thus ∑^ √arnn is bounded. Hence ∑^ √arnn converges.
Note: If we say ∑^ an converges faster than ∑^ bn, it means that
nlim→∞^ an bn
According the above exercise, we can construct a faster convergent se- ries from a known convergent one easily. It implies that there is no perfect tests to test all convergences of the series from a known con- vergent one.
- Prove that the Cauchy product of two absolutely convergent series con- verges absolutely.
Note: Given ∑^ an and ∑^ bn, we put cn = ∑nk=0 akbn−k and call ∑^ cn the Cauchy product of the two given series.
Proof: Put An = ∑nk=0 |ak|, Bn = ∑nk=0 |bk|, Cn = ∑nk=0 |ck|. Then
Cn = |a 0 b 0 | + |a 0 b 1 + a 1 b 0 | + ... + |a 0 bn + a 1 bn− 1 + ... + anb 0 | ≤ |a 0 ||b 0 | + (|a 0 ||b 1 | + |a 1 ||b 0 |) + ... +(|a 0 ||bn| + |a 1 ||bn− 1 | + ... + |an||b 0 |) = |a 0 |Bn + |a 1 |Bn− 1 + ... + |an|B 0 ≤ |a 0 |Bn + |a 1 |Bn + ... + |an|Bn = (|a 0 | + |a 1 | + ... + |an|)Bn = AnBn ≤ AB
where A = lim An and B = lim Bn. Hence {Cn} is bounded. Note that {Cn} is increasing, and thus Cn is a convergent sequence, that is, the Cauchy product of two absolutely convergent series converges absolutely.
- If {an} is a complex sequence, define its arithmetic means σn by
σn = s^0 +^ s n^1 ++ 1^ ... +^ sn(n = 0, 1 , 2 , ...).
(a) If lim sn = s, prove that lim σn = s. (b) Construct a sequence {sn} which does not converge, although lim σn =
(c) Can it happen that sn > 0 for all n and that lim sup sn = ∞, although lim σn = 0?
Choose M > 0 such that |tn| ≤ M for all n. Given > 0, choose N so that n > N implies |tn| < . Taking n > N in τn = (t 0 + t 1 + ... + tn)/(n + 1), and then
|τn| ≤ |t^0 |^ + n^ ... + 1^ + |tN^ |+ |tN^ +1 n^ + + 1^ ...^ +^ |tn
< (Nn^ + 1) + 1M + .
Hence, lim supn→∞ |τn| ≤ . Since is arbitrary, it follows that limn→∞ |τn| = 0, that is, lim σn = s.
Proof of (b): Let sn = (−1)n. Hence |σn| ≤ 1 /(n + 1), that is, lim σn = 0. However, lim sn does not exists.
Proof of (c): Let
sn =
1 , n = 0, n^1 /^4 + n−^1 , n = k^2 for some integer k, n−^1 , otherwise.
It is obvious that sn > 0 and lim sup sn = ∞. Also,
s 0 + ... + sn = 1 + nn−^1 +
n
⌋ n^1 /^4 = 2 +
n
⌋ n^1 /^4.
That is,
σn = (^) n + 1^2 + b
nc n^1 /^4 n + 1 The first term 2/(n + 1) → 0 as n → ∞. Note that
0 ≤ b
nc n^1 /^4 n + 1 < n
1 / (^2) n 1 / (^4) n− (^1) = n− 1 / (^4).
It implies that the last term → 0. Hence, lim σn = 0.
Proof of (d):
∑^ n k=
kak =
∑^ n k=
k(sk − sk− 1 ) =
∑^ n k=
ksk −
∑^ n k=
ksk− 1
=
∑^ n k=
ksk −
n∑− 1 k=
(k + 1)sk
= nsn +
n∑− 1 k=
ksk −
n∑− 1 k=
(k + 1)sk − s 0
= nsn −
n∑− 1 k=
sk − s 0 = (n + 1)sn −
∑^ n k=
sk = (n + 1)(sn − σn).
That is,
sn − σn = (^) n + 1^1
∑^ n k=
kak.
Note that {nan} is a complex sequence. By (a),
nlim→∞
n + 1
∑^ n k=
kak
) = limn→∞ nan = 0.
Also, lim σn = σ. Hence by the previous equation, lim s = σ.
Proof of (e): If m < n, then
∑^ n i=m+
(sn − si) + (m + 1)(σn − σm)
= (n − m)sn −
∑^ n i=m+
si + (m + 1)(σn − σm)
= (n − m)sn −
( (^) ∑n i=
si −
∑^ m i=
si
)
- (m + 1)(σn − σm) = (n − m)sn − (n + 1)σn + (m + 1)σm + (m + 1)(σn − σm) = (n − m)sn − (n − m)σn.
- Fix a positive number α. Choose x 1 >
α, and define x 2 , x 3 , x 4 , ..., by the recursion formula
xn+1 =^1 2 (xn + α xn
(a) Prove that {xn} decreases monotonically and that lim xn = √α. (b) Put n = xn − √α, and show that
n+1 =
(^2) n 2 xn^ <^
^2 n 2 √α so that, setting β = 2√α,
n+1 < β( β^1 )^2 n (n = 1, 2 , 3 , ...).
(c) This is a good algorithm for computing square roots, since the recursion formula is simple and the convergence is extremely rapid. For example, if α = 3 and x 1 = 2, show that 1 /β < 101 and therefore 5 < 4 · 10 −^16 , 6 < 4 · 10 −^32.
Proof of (a):
xn − xn+1 = xn −
2 (xn^ +^
α xn^ ) = 12 (xn − (^) xα n
= 12 (x
(^2) n − α xn^ )
0 since xn > α. Hence {xn} decreases monotonically. Also, {xn} is bounded by 0; thus {xn} converges. Let lim xn = x. Hence
lim xn+1 = lim^1 2 (xn + α xn ) ⇔ x =^1 2 (x + α x
⇔ x^2 = α.
Note that xn > 0 for all n. Thus x =
α. lim xn =
α. Proof of (b):
xn+1 =^1 2 (xn + α xn
⇒ xn+1 −
α =^12 (xn + (^) xα n
α
⇒ xn+1 −
α =^12 x
(^2) n − 2 xn^ √α + α xn ⇒ xn+1 −
α = (xn^ −
√α) 2 2 xn ⇒ n+1 =
(^2) n 2 xn^ <^
^2 n 2 √α. Hence n+1 < β( β^1 )^2 n where β = 2√α by induction. Proof of (c): 1 β =
Thus
5 < β( β^1 )^24 < 2
3 · 10 −^16 < 4 · 10 −^16 ,
6 < β( β^1 )^25 < 2
3 · 10 −^32 < 4 · 10 −^32.
Note: It is an application of Newton’s method. Let f (x) = x^2 − α in Exercise 5.25.
- Let X be a metric space. (a) Call two Cauchy sequences {pn}, {qn} in X equivalent if nlim→∞ d(pn, qn) = 0. Prove that this is an equivalence relation.
(b) Let X∗^ be the set of all equivalence classes so obtained. If P ∈ X∗, Q ∈ X∗, {pn} ∈ P , {qn} ∈ Q, define 4 (P, Q) = limn→∞ d(pn, qn); by Exercise 23, this limit exists. Show that the number 4 (P, Q) is unchanged if {pn} and {qn} are replaced by equivalent sequences, and hence that 4 is a distance function in X∗.
(c) Prove that the resulting metric space X∗^ is complete. Proof of (a): Suppose there are three Cauchy sequences {pn}, {qn}, and {rn}. First, d(pn, pn) = 0 for all n. Hence, d(pn, pn) = 0 as n → ∞. Thus it is reflexive. Next, d(qn, pn) = d(pn, qn) → 0 as n → ∞. Thus it is symmetric. Finally, if d(pn, qn) → 0 as n → ∞ and if d(qn, rn) → 0 as n → ∞, d(pn, rn) ≤ d(pn, qn) + d(qn, rn) → 0 + 0 = 0 as n → ∞. Thus it is transitive. Hence this is an equivalence relation. Proof of (b): Proof of (c): Let {Pn} be a Cauchy sequence in (X∗, 4 ). We wish to show that there is a point P ∈ X∗^ such that 4 (Pn, P ) → 0 as n → ∞. For each Pn, there is a Cauchy sequence in X, denoted {Qkn}, such that 4 (Pn, Qkn) → 0 as k → ∞. Let n > 0 be a sequence tending to 0 as n → ∞. From the double sequence {Qkn} we can extract a subsequence Q′ n such that 4 (Pn, Q′ n) < n for all n. From the triangle inequality, it follows that 4 (Q′ n, Q′ m) ≤ 4(Q′ n, Pn) + 4 (Pn, Pm) + 4 (Pm, Q′ m). (1)
Since {Pn} is a Cauchy sequence, given > 0, there is an N > 0 such that 4 (Pn, Pm) < for m, n > N. We choose m and n so large that m < , n < . Thus (1) shows that {Q′ n} is a Cauchy sequence in X. Let P be the corresponding equivalence class in S. Since
4 (P, Pn) ≤ 4(P, Q′ n) + 4 (Q′ n, Pn) < 2
for n > N , we conclude that Pn → P as n → ∞. That is, X∗^ is complete.
