Numerical Sequences and Series

Numerical Sequences and Series Written by Men-Gen Tsai email: b89902089@ntu.edu.tw. Prove that the convergence of {s n } implies convergence of { s n }. Is the converse true? Solution: Since {s n } is convergent, for any ɛ > 0, there exists N such that s n s < ɛ whenever n N. By Exercise.3 I know that s n s s n s. Thus, s n s < ɛ, that is, {s n } is convergent. The converse is not true. Consider s n = ( ) n. 2. Calculate lim n ( n 2 + n n). Solution: n2 + n n = = n n2 + n + n /n + + 2 as n. 3. If s n = 2 and s n+ = 2 + s n (n =, 2, 3,...), prove that {s n } converges, and that s n < 2 for n =, 2, 3,... Proof: First, I show that {s n } is strictly increasing. It is trivial that s 2 = 2 + s = 2 + 2 > 2 = s. Suppose s k > s k when

k < n. By the induction hypothesis, s n = > 2 + s n 2 + s n 2 = s n By the induction, {s n } is strictly increasing. Next, I show that {s n } is bounded by 2. Similarly, I apply the induction again. Hence {s n } is strictly increasing and bounded, that is, {s n } converges. 4. 5. 6. 7. Prove that the convergence of implies the convergence of if 0. Proof: By Cauchy s inequality, n k k n= n= n 2 k n= an n for all n N. Also, both and are convergent; thus k n 2 n= a an n n is bounded. Besides, 0 for all n. Hence is convergent. n n 8. 9. Find the radius of convergence of each of the following power series: (a) n 3 z n, (b) 2 n n! zn, (c) 2 n n 2 zn, (d) n 3 3 n zn. 2

Solution: (a) α n = (n 3 ) /n as n. Hence R = /α =. (b) α n = (2 n /n!) /n = 2/(n!) /n 0 as n. Hence R = +. (c) α n = (2 n /n 2 ) /n 2/ = 2 as n. Hence R = /α = /2. (d) α n = (n 3 /3 n ) /n /3 as n. Hence R = /α = 3. 0.. Suppose > 0, s n = a +... +, and diverges. (a) Prove that + diverges. (b) Prove that a N+ +... + a N+k s N s N+ s N+k and deduce that sn (c) Prove that diverges. s 2 n converges. and deduce that s 2 n (d) What can be said about Proof of (a): Note that s n s n s N+k + n and + n 2? + 0 + 0 0 as n. If + converges, then 0 as n. Thus for some ɛ = there is an N such that < whenever n N. Since an + converges, for any ɛ > 0 there is an N 2 such that a m + a m +... + + < ɛ 3

all n > m N 2. Take N = max(n, N 2 ). Thus for all n > m N. Thus a m ɛ > +... + + a m + > a m + +... + + = a m +... + 2 a m +... + < 2ɛ for all n > m N. It is a contradiction. Hence + diverges. Proof of (b): If sn a N+ s N+ +... + a N+k s N+k a N+ s N+k +... + a N+k s N+k = a N+ +... + a N+k s N+k = s N+k s N s N+k = s N s N+k converges, for any ɛ > 0 there exists N such that a m s m +... + s n < ɛ for all m, n whenever n > m N. Fix m = N and let n = N + k. Thus ɛ > a m +... + s m s n = a N +... + a N+k s N s N+k s N s N+k 4

for all k N. But s N+k as k since diverges and > 0. Take ɛ = /2 and we obtain a contradiction. Hence sn diverges. Proof of (c): for all n. Hence Note that s n Proof of (d): s n s n s 2 n k n=2 s 2 n s 2 n s 2 n s n s n s n s n = s n s n s n s n s n s n k ( ) n=2 s n s n = s s n. 0 as n since diverges. Hence s 2 n +n may converge or diverge, and a, that is, +n 2 converges. To see this, we put = /n. 2 /n diverges. Besides, if we put where p > and n 2, then + n = < = n(log n) p +n = 2n n(log n) 2p ((log n) p + ) 2n(log n) 3p converges. n +n = for large enough n. By Theorem 3.25 and Theorem 3.29, +n converges. Next, = + n 2 / + n 2 5

< n 2. for all. Note that n 2 converges, and thus +n 2 converges. 2. Suppose > 0 and converges. Put r n = a m. m=n (a) Prove that a m +... + > r n r m r n r m if m < n, and deduce that rn diverges. (b) Prove that and deduce that rn Proof of (a): if m < n. If rn rn < 2( r n r n+ ) converges. a m r m +... + r n > a m +... + r m = r m r n r m = r n r m converges, for any ɛ > 0 there exists N such that a m r m +... + r n < ɛ for all m, n whenever n > m N. Fix m = N. Thus a m r m +... + r n > r n r m = r n r N 6

for all n > N. But r n 0 as n ; thus am r m +...+ an r n as n. If we take ɛ = /2, we will get a contradiction. Proof of (b): Note that r n+ < r n r n+ < r n r n + r n+ < 2 r n rn + r n+ < 2 rn ( r n rn + r n+ r n+ ) < 2( r n r n+ ) rn r n r n+ rn < 2( r n r n+ ) rn < 2( r n r n+ ) since > 0 for all n. Hence, k k < 2( r n r n+ ) n= rn n= = 2( r r k+ ) Note that r n 0 as n. Thus rn converges. is bounded. Hence rn Note: If we say converges faster than b n, it means that lim = 0. n b n According the above exercise, we can construct a faster convergent series from a known convergent one easily. It implies that there is no perfect tests to test all convergences of the series from a known convergent one. 7

3. Prove that the Cauchy product of two absolutely convergent series converges absolutely. Note: Given and b n, we put c n = n k=0 a k b n k and call c n the Cauchy product of the two given series. Proof: Put A n = n k=0 a k, B n = n k=0 b k, C n = n k=0 c k. Then C n = a 0 b 0 + a 0 b + a b 0 +... + a 0 b n + a b n +... + b 0 a 0 b 0 + ( a 0 b + a b 0 ) +... +( a 0 b n + a b n +... + b 0 ) = a 0 B n + a B n +... + B 0 a 0 B n + a B n +... + B n = ( a 0 + a +... + )B n = A n B n AB where A = lim A n and B = lim B n. Hence {C n } is bounded. Note that {C n } is increasing, and thus C n is a convergent sequence, that is, the Cauchy product of two absolutely convergent series converges absolutely. 4. If { } is a complex sequence, define its arithmetic means σ n by σ n = s 0 + s +... + s n (n = 0,, 2,...). n + (a) If lim s n = s, prove that lim σ n = s. (b) Construct a sequence {s n } which does not converge, although lim σ n = 0. (c) Can it happen that s n > 0 for all n and that lim sup s n =, although lim σ n = 0? 8

(d) Put = s n s n, for n. Show that s n σ n = n + ka k. k= Assume that lim(n ) = 0 and that {σ n } converges. Prove that {s n } converges. [This gives a converse of (a), but under the additional assumption that n 0.] (e) Derive the last conclusion from a weaker hypothesis: Assume M <, n M for all n, and lim σ n = σ. Prove that lim s n = σ, by completing the following outline: If m < n, then For these i, s n σ n = m + n m (σ n σ m ) + n m i=m+ (s n s i ). s n s i (n i)m i + (n m )M. m + 2 Fix ɛ > 0 and associate with each n the integer m that satisfies m n ɛ + ɛ < m +. Then (m + )/(n m) /ɛ and s n s i < Mɛ. Hence Since ɛ was arbitrary, lim s n = σ. lim sup s n σ Mɛ. n Proof of (a): The proof is straightforward. Let t n = s n s, τ n = σ n s. (Or you may suppose that s = 0.) Then τ n = t 0 + t +... + t n. n + 9

Choose M > 0 such that t n M for all n. Given ɛ > 0, choose N so that n > N implies t n < ɛ. Taking n > N in τ n = (t 0 + t +... + t n )/(n + ), and then τ n t 0 +... + t N + t N+ +... + t n n + n + (N + )M < + ɛ. n + Hence, lim sup n τ n ɛ. Since ɛ is arbitrary, it follows that lim n τ n = 0, that is, lim σ n = s. Proof of (b): Let s n = ( ) n. Hence σ n /(n + ), that is, lim σ n = 0. However, lim s n does not exists. Proof of (c): Let, n = 0, s n = n /4 + n, n = k 2 for some integer k, n, otherwise. It is obvious that s n > 0 and lim sup s n =. Also, That is, s 0 +... + s n = + nn + n n /4 = 2 + n n /4. σ n = 2 n + + n n /4 n + The first term 2/(n + ) 0 as n. Note that 0 n n /4 n + < n /2 n /4 n = n /4. It implies that the last term 0. Hence, lim σ n = 0. 0

Proof of (d): ka k = k(s k s k ) = ks k ks k k= k= k= k= n = ks k (k + )s k k= k=0 n n = ns n + ks k (k + )s k s 0 k= k= n = ns n s k s 0 = (n + )s n s k k= k=0 = (n + )(s n σ n ). That is, s n σ n = n + ka k. k= Note that {n } is a complex sequence. By (a), ( lim n n + k= ka k ) = lim n n = 0. Also, lim σ n = σ. Hence by the previous equation, lim s = σ. Proof of (e): If m < n, then (s n s i ) + (m + )(σ n σ m ) i=m+ = (n m)s n s i + (m + )(σ n σ m ) i=m+ ( n m ) = (n m)s n s i s i + (m + )(σ n σ m ) i=0 i=0 = (n m)s n (n + )σ n + (m + )σ m + (m + )(σ n σ m ) = (n m)s n (n m)σ n.

Hence, s n σ n = m + n m (σ n σ m ) + n m i=m+ (s n s i ). For these i, recall = s n s n and n M for all n, s n s i = k=i+ a k k=i+ a k k=i+ (n (m + ))M = (m + ) + M i + (n m )M. m + 2 = (n i)m i + Fix ɛ > 0 and associate with each n the integer m that satisfies Thus or Hence, m n ɛ + ɛ < m +. n m m + ɛ and n m m + 2 m + n m ɛ and < ɛ, s n s i < Mɛ. s n σ σ n σ + ɛ ( σ n σ + σ m σ ) + Mɛ. Let n and thus m too, and thus Since ɛ was arbitrary, lim s n = σ. lim sup s n σ Mɛ. n 5. 2

6. Fix a positive number α. Choose x > α, and define x 2, x 3, x 4,..., by the recursion formula x n+ = 2 (x n + α x n ). (a) Prove that {x n } decreases monotonically and that lim x n = α. (b) Put ɛ n = x n α, and show that so that, setting β = 2 α, ɛ n+ = ɛ2 n 2x n < ɛ2 n 2 α ɛ n+ < β( ɛ β )2n (n =, 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 = 2, show that ɛ /β < 0 ɛ 5 < 4 0 6, ɛ 6 < 4 0 32. and therefore Proof of (a): x n x n+ = x n 2 (x n + α x n ) = 2 (x n α x n ) = 2 (x2 n α x n ) > 0 since x n > α. Hence {x n } decreases monotonically. Also, {x n } is bounded by 0; thus {x n } converges. Let lim x n = x. Hence lim x n+ = lim 2 (x n + α x n ) x = 2 (x + α x ) 3 x 2 = α.

Note that x n > 0 for all n. Thus x = α. lim x n = α. Proof of (b): x n+ = 2 (x n + α x n ) x n+ α = 2 (x n + α x n ) α x n+ α = x 2 n 2x n α + α 2 x n x n+ α = (x n α) 2 2x n ɛ n+ = ɛ2 n 2x n < ɛ2 n 2 α. Hence where β = 2 α by induction. Proof of (c): ɛ n+ < β( ɛ β )2n ɛ β = 2 3 2 3 = 2 3(2 + 3) = 6 + 4 3 < 0. Thus ɛ 5 < β( ɛ β )24 < 2 3 0 6 < 4 0 6, ɛ 6 < β( ɛ β )25 < 2 3 0 32 < 4 0 32. Note: It is an application of Newton s method. Let f(x) = x 2 α in Exercise 5.25. 7. 8. 4

9. 20. 2. 22. Suppose X is a complete metric space, and {G n } is a sequence of dense open subsets of X. Prove Baire s theorem, namely, that G n is not empty. (In fact, it is dense in X.) Hint: Find a shrinking sequence of neighborhoods E n such that E n G n, and apply Exercise 2. Proof: I ve proved it in Chapter 2 Exercise 30. 23. Suppose {p n } and {q n } are Cauchy sequences in a metric space X. Show that the sequence {d(p n, q n )} converges. Hint: For any m, n, d(p n, q n ) d(p n, p m ) + d(p m, q m ) + d(q m, q n ); it follows that d(p n, q n ) d(p m, q m ) is small if m and n are large. Proof: For any ɛ > 0, there exists N such that d(p n, p m ) < ɛ and d(q m, q n ) < ɛ whenever m, n N. Note that d(p n, q n ) d(p n, p m ) + d(p m, q m ) + d(q m, q n ). It follows that d(p n, q n ) d(p m, q m ) d(p n, p m ) + d(q m, q n ) < 2ɛ. Thus {d(p n, q n )} is a Cauchy sequence in X. Hence {d(p n, q n )} converges. 5

24. Let X be a metric space. (a) Call two Cauchy sequences {p n }, {q n } in X equivalent if lim d(p n, q n ) = 0. n Prove that this is an equivalence relation. (b) Let X be the set of all equivalence classes so obtained. If P X, Q X, {p n } P, {q n } Q, define by Exercise 23, this limit exists. (P, Q) = lim n d(p n, q n ); Show that the number (P, Q) is unchanged if {p n } and {q n } are replaced by equivalent sequences, and hence that 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 {p n }, {q n }, and {r n }. First, d(p n, p n ) = 0 for all n. Hence, d(p n, p n ) = 0 as n. Thus it is reflexive. Next, d(q n, p n ) = d(p n, q n ) 0 as n. Thus it is symmetric. Finally, if d(p n, q n ) 0 as n and if d(q n, r n ) 0 as n, d(p n, r n ) d(p n, q n ) + d(q n, r n ) 0 + 0 = 0 as n. Thus it is transitive. Hence this is an equivalence relation. Proof of (b): Proof of (c): Let {P n } be a Cauchy sequence in (X, ). We wish to show that there is a point P X such that (P n, P ) 0 as n. For each P n, there is a Cauchy sequence in X, denoted {Q k n}, such that (P n, Q k n) 0 as k. Let ɛ n > 0 be a sequence tending to 0 as n. From the double sequence {Q k n} we can extract a subsequence Q n such that (P n, Q n) < ɛ n for all n. From the triangle inequality, it follows that (Q n, Q m) (Q n, P n ) + (P n, P m ) + (P m, Q m). () 6

Since {P n } is a Cauchy sequence, given ɛ > 0, there is an N > 0 such that (P n, P m ) < ɛ for m, n > N. We choose m and n so large that ɛ m < ɛ, ɛ n < ɛ. Thus () shows that {Q n} is a Cauchy sequence in X. Let P be the corresponding equivalence class in S. Since (P, P n ) (P, Q n) + (Q n, P n ) < 2ɛ for n > N, we conclude that P n P as n. That is, X is complete. 25. 7