MA541 : Real Analysis Tutorial and Practice Problems - 1 Hints and Solutions 1. Suppose that S is a nonempty subset of real numbers that is bounded (i.e. bounded above as well as below). Prove that inf S ≤ sup S. What can you say if inf S = sup S? Solution: Second Part: inf S = sup S iﬀ S is a singleton. 2. Is the set {x ∈ R | x2 < x} bounded above (below)? If so, ﬁnd its supremum (inﬁmum). Solution: Hint. {x ∈ R | x2 < x} = (0, 1). 3. Suppose n is a positive integer and a ∈ R, a ≥ 0. Show that there is a unique b ≥ 0 in R such that bn = a. Solution: Hint. Consider the set S = {x ∈ R : x ≥ 0, xn ≤ a}. Show that S is nonempty and bounded above. The supremum b of S satisﬁes bn = a (show that bn < a and bn > a are not possible). Finally show that that if c ≥ 0 and cn = a, then c = b. 4. Show that between any two distinct real numbers there is a rational number as well as an irrational number. Solution: (Refer to any text book.) 5. Prove Well-Ordering Principle of natural numbers: Every nonempty subset of positive integers has a minimum. Solution: Let S be a nonempty subset of positive integers. Then S is bounded below by 0, and so has an inﬁmum, say c. Claim: c is the least element of S. Since c + 1 is not a lower bound of S, there is m ∈ S such that m < c + 1. Now, for any positive integer n < m we have n ≤ m − 1 < c and therefore, n∈ / S. Consequently, m is the least element of S. (Note that this also means that c = m.) 1 of 5 BK Sarma, 05 Aug 2014, 18:42 6. Suppose that a real number a has the property that 1 n ≥ a for all n ∈ N. Show that a ≤ 0. Solution: Hint. Follows from Archimedean Principle. 7. Let S and T be nonempty sets of real numbers such that every real number is in S or T and if s ∈ S and t ∈ T , then s < t. Prove that there is a unique real number β such that every real number less than β is in S and every real number greater than β is in T . (A decomposition of the reals into two sets with these properties is a Dedekind cut. This is known as Dedekind’s theorem.) Solution: Done in the Tutorial class. 8. Suppose that a sequence (xn ) converges to x and xn ≥ 0 for all n. Show that x ≥ 0. Solution: Hint. Show that if x < 0, then xn < 0 for inﬁnitely many values of n. 9. A subset S of R is closed if every convergent sequence in S has its limit in S. Which of the following sets are closed: [a, b), [a, b], Q, R \ Q? Justify your answer. Solution: [a, b] is closed: If xn → x and a ≤ xn ≤ b for each n, then a ≤ x ≤ b. [a, b) is not closed: If xn = b − b−a / [a, b). n , then xn ∈ [a, b) and xn → b ∈ Q and R \ Q are not closed: Use density property. 10. Let (xn ), (yn ), (zn ) be sequences such that xn ≤ yn ≤ zn for all n ∈ N. If both (xn ) and (zn ) converges to the same limit c, then show that (yn ) converges to c. [This result is known as the Sandwich Theorem.] Solution: Refer to any text book. ) ( xn+1 converges to L. Show that (xn ) (i) converges to 0, if 11. Let (xn ) be such that xn ̸= 0 and xn L < 1, and (ii) diverges, if L > 1. Show that (xn ) can converge or diverge, if L = 1. Solution: Done in the Tutorial class. 2 of 5 BK Sarma, 05 Aug 2014, 18:42 12. Let a, b ∈ R, and let (xn ) be deﬁned as follows: x1 = a, x2 = b, and xn = 12 (xn−2 + xn−1 ) for n ≥ 3. Prove that (xn ) is convergent and ﬁnd its limit. Solution: For n ≥ 3, we have xn = 12 (xn−2 +xn−1 ) and therefore xn −xn−1 = − 12 (xn−1 +xn−2 ), i.e., |xn − xn−1 | = 12 |xn−1 + xn−2 |. Therefore, by the Result given below, (xn ) is a Cauchy sequence and so convergent. Moreover, xn − x1 = (xn − xn−1 ) + (xn−1 − xn−2 ) + · · · + (x2 − x1 ) ] [( ) ( )n−3 1 1 n−2 + − + · · · + 1 (b − a) = − 2 2 [ ( )n−1 ] 2 1 = (b − a). 1− − 3 2 Taking limits we get lim xn − a = 23 (b − a), i.e., lim xn = a + 32 (b − a) 13. Let (xn ) be a sequence and yn = n1 (x1 + x2 + · · · + xn ). Show that if (xn ) is convergent, then (yn ) is convergent. Is the converse true? Solution: Suppose xn → a. Let ϵ > 0 be given. There exists N ∈ N such that |xn − a| < ϵ/2 for all n ≥ N . Now, for n ≥ N 1 |yn − a| = (x1 + · · · + xn ) − a n [ ] ≤ |x1 − a| + |x2 a| + · · · + |xn − a| = N n 1 ∑ 1∑ |xi − a| + |xi − a| n n = (n − N ) ϵ K + . n n 2 i=1 i=N +1 ∑ where K = N i=1 |xi − a|. Choose positive integer M ≥ N such that K/M < ϵ/2. Then for n ≥ M we have |yn − a| < ϵ. The converse is not true. Take for example, xn = (−1)n . 14. Prove or disprove: If (xn ) converges to x, and (xn ) has a subsequence (xnk ) such that (−1)k xnk ≥ 0 for all k, then x = 0. ( ) Solution: Suppose, if possible, x > 0. There exists m ∈ N such that xn ∈ x2 , 3x 2 for all n ≥ m (Note: we took ϵ = x/2.) In particular, xn > 0 for all n ≥ m. Now, nm , nm+1 ≥ m and therefore xnm , xnm+1 are positive. This implies that (−1)m xnm ≥ 0, (−1)m+1 xnm+1 ≥ 0 simultaneously cannot hold, since one of them is negative. Thus, x ≤ 0. Similarly, it can be shown that x < 0 is not possible. 3 of 5 BK Sarma, 05 Aug 2014, 18:42 15. Examine whether the sequences (xn ) converge, where xn is given below. Find also the limits, if exist. Here, a, b are real numbers. 1 n (iv) sin3 n ( )n 1 + n1 (vii) nn (i) 1 (iii) an n! (an + bn ) n , where a, b > 0 (vi) 1 n+1 3·5·7···(2n−1) 2·5·8···(3n−1) (ix) 1 n2 (ii) an (v) (viii) Solution: (i) − n1 ≤ 1 sin3 n n ( + 1 n+2 + ··· + 1 n+n ) [a] + [2a] + · · · + [na] ≤ n1 . Now use Sandwich Theorem. (ii) xn = an . If a = 0 or 1, then clearly (xn ) converges. If a = −1, then (xn ) = ((−1)n ) diverges. Case: |a| > 1. Let |a| = 1 + h. Then n(n − 1) 2 h + · · · + hn > nh. 2 Thus, (xn ) is not bounded, and so not convergent. 1 Case: 0 < a < 1. Then 1/a > 1. Put 1/a = 1 + h, i.e., a = 1+h . Thus, |an | = (1 + h)n = 1 + nh + 0 < x n = an = 1 1 1 = ≤ . n n (1 + h) 1 + nh + · · · + h 1 + nh Use Sandwich Theorem to conclude that xn → 0. Case: −1 < a < 0. Then |an | → 0 and therefore an → 0. |a| (iii) xxn+1 = n+1 → 0 < 1. Thus xn → 0. n ( )n (iv) Show that 2 < xn = 1 + n1 < 3 and (xn ) is monotonically increasing. Thus the sequence converges. (lim xn is denoted by e.) 1 1 1 (v) Assume a ≤ b. Then b ≤ xn = (an + bn ) n ≤ (bn + bn ) n = 2 n b. Use Sandwich Theorem to conclude xn → b. Thus, xn → max{a, b}. 1 1 n 1 + n+2 + · · · + n+n ≤ n+1 < 1, the sequence (xn ) is bounded. Moreover, (vi) Since 0 < xn = n+1 1 1 1 xn+1 − xn = 2n+1 + 2n+2 − n+1 > 0, i.e., (xn ) is monotonically increasing. Hence (xn ) converges. In fact, [ ] ∫ 1 1 1 1 1 dx xn = + + ··· + → = ln 2. n 1 + 1/n 1 + 2/n 1 + n/n 0 1+x 1 1 (vii) xn = n n . We have xn > 1 for n ≥ 2. Put hn = xn − 1. Then n n = 1 + hn , i.e., n = (1 + hn )n = 1 + nhn + n(n − 1) 2 n(n − 1) 2 hn + · · · > hn . 2 2 √ 2 2 i.e., 0 ≤ hn ≤ √n−1 . Using Sandwich Theorem, we Therefore, for n ≥ 2 we have 0 ≤ h2n ≤ n−1 get hn → 0, i.e. xn → 1. 2n+1 (viii) Note that xxn+1 = 3n+1 → 2/3 < 1. Hence xn → 0. n ( ) (ix) xn = n12 [a] + [2a] + · · · + [na] . Note that for any real b, b − 1 < [b] ≤ b. Thus, ) ) 1( 1( (a − 1) + (2a − 1) + · · · + (na − 1) < x ≤ a + 2a + · · · + na . n n2 n2 Simplify this to get 1 a n+1 a n+1 − + · < xn ≤ · . n 2 n 2 n Use Sandwich Theorem and get xn → a2 . 4 of 5 BK Sarma, 05 Aug 2014, 18:42 Result. Suppose (xn ) be a real sequence such that |xn+1 − xn | ≤ c|xn − xn−1 | for all n. (0.1) If c < 1, then (xn ) is a Cauchy sequence and therefore convergent. Proof. If x2 = x1 , then xn = x1 for all n, and the result follows. Suppose x2 ̸= x1 . We have for all n |xn+1 − xn | ≤ c|xn − xn−1 | ≤ · · · ≤ cn−1 |x2 − x1 |. Moreover, |xn+k − xn | ≤ |xn+k − xn+k−1 | + · · · + |xn−1 − xn | ≤ (ck−1 + cn−2 + · · · + c + 1)|xn−1 − xn | 1 − ck = |xn1 − xn | 1−c 1 ≤ cn−1 |x2 − x1 |. 1−c c(1 − c) ϵ for all |x2 − x1 | n ≥ N . Then for any n ≥ N, k ≥ 1 we have |xn+k − xn | < ϵ. Thus (xn ) is a Cauchy sequence. Let ϵ > 0 be given. Since 0 ≤ c < 1, we have cn → 0. Choose N ∈ N such that cn < 5 of 5 BK Sarma, 05 Aug 2014, 18:42

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