MA8S Tutorial Solution Zhang Liyang, Xiong Xi, Sun Bo Tutorial on limit superior and inferior (). Given lim(a n ) and (b n ) is bounded, show that lim a nb n = lim a n lim b n n n n Proof. Let u := lim n a n b n, a := lim(a n ), b := lim n b n. When a >, let x be a cluster point of (a n b n ). Then there is a convergent subsequence (a nk b nk ) that converges to x. Since (b n ) is bounded, (b nk ) is also bounded, by the Bolzano-Weierstrass Theorem, (b nk ) has a convergent subsequence (b nkl ), say lim(b nkl ) = p. Since (a n ) is convergent, (a nkl ) also converges to a, so x = lim(a nk b nk ) = ap, since u = lim n a n b n, so ap u, p u, since b = lim a n b n, we have b u, so ab u. a Let q be a cluster point of (b n ). Then there is a convergent subsequence (b nk ), lim(b nk ) = q, since (a n ) is convergent, (a nk ) is convergent, (a nk b nk ) is convergent, then lim(a nk b nk ) = aq, q lim n b n, so aq ab, so ab is an upper bound for S(a n b n ), since u = lim n a n b n, then ab u, therefore, we have lim n a n b n = lim n a n lim n b n When a =, since (b n ) is bounded, M > such that b n M, n N. Let ɛ > be given. Then since (a n ) is convergent, there exists K N such that if n K, then a n < ɛ M, that is, a n < ɛ M, so if n K, a nb n < a n b n < ɛ, so lim(a n b n ) =. Therefore, lim n a n b n = = lim(a n )lim n b n. (3). Let a n > for n N. If lim n a n+ /a n = r,show that {a n } n= converges if r <. What if r? Proof. Case(i): r <. Let ɛ = r. Since lim a n+ a n = r and ɛ >, a n N such that when n > n, n+ a n < r + ɛ = +r <. Then +r < a n+n < a n+n < < ( +r )n a n. Since lim( +r )n+ a n =, lim a n+n =. Hence lim a n =
Case(ii) r > : inconclusive. While the sequence (r n ) diverges if r >, the sequence a n = r/n when n is odd and /n if n is even has lim n a n+ /a n = r and the sequence (a n ) converges to. Case(iii) r = : inconclusive. The sequence (,,, ) converges, while (,,,,,,, 3,,,, 3,,, )diverges. 3 3 3 4 4 4 4 4 (4). Show that lim ( sin n xdx) /n = n. Proof. When x π 6, we have sin x Thus ( sin n xdx) /n (π 6 )/n Since lim ( π 6 )/n = lim( π 6 )/n =, therefore, we have lim ( n On the other hand, since sin n xdx) /n lim n (π 6 )/n =. lim sin x = x π/6, for < ɛ <, there exist < δ < π such that ɛ < sin x <. 6 Thus,( sin n xdx) /n ( π/6 δ sinn xdx) /n. Therefore, lim n ( Since ɛ can be arbitrary small, therefore, sin n xdx) /n lim n δ n ( ɛ) = ɛ. lim n ( sin n xdx) /n.
Thus, lim ( sin n xdx) /n = n. (6). Show that the set of clutser points of the sequence (( ) n ) is {, }. Proof. Clearly there are subsequences (( ) n ) = () and (( ) n ) = ( ) that converge to and - respectively. Now we show there is no other cluster point other than and. If there is a subsequence (a nk ) that converges to l, l and l. Let ɛ = min{ l, + l }. Clearly ɛ >. But for all N N, there exists k > N, a nk = or a nk =, a nk l min{ l, + l } = ɛ, which is a contradiction. 3.6.9 Let (x n ) and (y n ) be sequences of positive numbers such that lim(x n /y n ) = + (a) Show that if lim(y n ) = +, then lim(y n ) = +. Proof. (a) Note that (x n ) and (y n ) are all positive sequences. Because lim(x n /y n ) = +, there exist N N, such that for all n > N, x n /y n >, i.e., x n > y n. For any M, as lim(y n ) = +, there exist N N, such that when n > N, y n > M. Hence when n > max{n, N }, x n > y n > M. Hence lim(x n ) = +. (b) Show that if (y n ) is bounded, then lim(x n ) =. Note that (x n ) and (y n ) are all positive sequences. As x n is bounded, x n M for all n N. For all ɛ >, because lim(x n /y n ) = +, there exists N N, such that for all n > N, x n yn Hence lim(y n ) =. > M ɛ, which implies y n = y n < ɛx n M ɛ. 3
3.7. Let a n be a given series and let b n be the series in which the terms are the same and in the same order as in a n except that the terms for which a n = have been omitted. Show that a n converges to A if and only if bn converges to A. Proof. Since the terms for which a n = has been omitted, the sequence of initial sums ( k n= b n) is a subsequence of ( k n= a n). Hence if a n converges to A, then b n converges to A. Conversely, if b n converges to A. For all ɛ >, there exists N N, such that when n > N, n k= b k A < ɛ. According to the definition of (b n ), there exists N N, such that N k= b k = N k= a k. When n > N, n k= a k A = n k= b k A < ɛ, for an n, n > N. Hence a n converges to A. 3.7. Show that the convergence of a series is not affected by changing a finite number of its terms.(of course, the value of the sum may be changed.) Proof. Let (a n ) denote the series. Let S = {n : a n is changed}. As S is finite, there exists N N, such that when n > N, a n is not changed. For all ɛ >, as (a n ) converges, there exists N N, such that when m > n > N, m n+ a n < ɛ. When m > n > max{n, N }, m n+ b n = m n+ a n < ɛ. Hence (b n ) also converges. 3.7.3 By using partial fractions, show that (a) n= = (n+)(n+) 4
Solution: Since =,we have (n+)(n+) n+ n+ s n = + + + = n+. Hence lim(s 3 n+ n+ n+ n) = (b) n= =, if α > (α+n)(α+n+) α Solution: (α+n)(α+n+) = α+n α+n+. Hence s n = α α + + α + + + α + α + n α + n + = α α + n +. Thus lim(s n ) = α. 3.7.4 If x n and y n are convergent, show that x n + y n is convergent. Proof. Let s n denotes the nth partial sum of x n and t n denotes the nth partial sum of y n. Then s n + t n is the nth partial sum of (x n + y n ) Assume lim(s n ) = s and lim(t n ) = t. By limit theorems, lim(s n + t n ) = s + t, thus (x n + y n ) is convergent. 3.7.5 Can you give an example of a convergent series x n and a divergent series yn such that (x n + y n ) is convergent? Explain. Solution: No. If x n converges, then ( x n ) converges. By 3.7.4, y n = xn + y n x n is convergent, resulting in a contradiction. Alternative solution: Assume x n and x n + y n are both convergent. Given ɛ >, by definition, there exist M (N) such that x n+ +...+x m < ɛ for all m > n M. There exist M (N) such that (x n+ + y n+ )... + (x m + y m ) < ɛ for all m > n M. Let M = maxm, M, then for all m > n M,we have y n+ +... + y m = (x n+ + y n+ +... + (x m + y m ) (x n+ +...x m ) 5
(x n+ + y n+ )... + (x m + y m ) + x n+ +... + x m < ɛ + ɛ yn is convergent. = ɛ. Therefor Thus, we can not find a convergent x n and a divergent y n such that (xn + y n ) is convergent. 3.7.6 (a) Show that the series n= cos n is divergent. Since (cos n) does not converge to, the series diverges. (b) Show that the series n= (cos n)/n is convergent. Since cos n n n and n is convergent, by the Comparison Test, n= (cos n)/n is convergent. So n= (cos n)/n is absolutely convergent, hence n= converges. (cos n)/n 3.7.8 If a n with a n > is convergent, then is a n always convergent? Either prove it or give a counterexample. Solution: Since a n, converges, there exists M > such that a n < M, since a n >, we have < a n < M, < a n < Ma n. Since Ma n converges, a n converges by the comparison test. Alternative solution: Let ɛ > be given, since a n is convergent, there exists M( ɛ) N such that if m > n M( ɛ), then < a n+ + a n+ + + a m < ɛ, so we have a n+ + a n+ + + a m (a n+ + a n+ + + a m ) < ɛ so a n converges. Question(C) Show that if we define the positive real numbers as above, then the set of real number satisfies properties..5. 6
() We show that if [(a n )] P, [(b n )] P, then [(a n + b n )] P, [(a n b n )] P. Since [(a n )] P, (a n) [(a n )] with a n > r, for all n for some positive rational number r; similarly, (b n) [(b n )] with b n > r for all n for some positive rational number r. Thus, a n + b n > r + r >, a nb n > r r > for all n N. Therefore, [(a n + b n )] = [(a n + b n)] P, [(a n b n )] = [(a nb n)] P. () We show that for any [(a n )], exactly one of the following statement is true : (i)[(a n )] P (ii) [(a n )] = [()] (iii) [(a n )] P. Case(i): If [(a n )] =, then for all r > and for all (a n) [(a n )], there exists K N such that a m < r m K, that is, a m < r and a m < r for all m K. Thus [(a n )] / P and [(a n )] / P. Case(ii): if [(a n )], we show that exactly one of the following is true: (i) [(a n )] P (ii) [(a n )] P. To see this, recall that (a n ) () implies there exists r Q + such that for all K N, there exists N K K such that a NK r. Thus, there exists a subsequence (a nk ) of (a n ) with a nk r for all k. Hence it will have either a subsequence with all terms r or a subsequence with all terms r. However, as (a n ) is a Cauchy sequence, at most one of them holds. In the first case where all terms r > r/, we have [(a n )] P, and in the second case, clearly [(a n )] P. 7