Problem Set 2 with Solutions | First Year Interest Group in Seminar | N 1, Assignments of Health sciences

Download Problem Set 2 with Solutions | First Year Interest Group in Seminar | N 1 and more Assignments Health sciences in PDF only on Docsity! M365C (Fall 2006) Problem Set 2 Due: Friday September 15, 2006 1 1. Let {xn}n∈N be a sequence of non-negative real numbers such that lim n→∞ xn = 0. Prove that the sequence {√xn}n∈N also converges, and lim n→∞ √ xn = 0. 2. Prove that if {xn}n∈N is an unbounded nondecreasing sequence of real numbers, then lim n→∞ xn = +∞. 3. Let {xn}n∈N be a sequence of real numbers. Prove that if lim sup n→∞ xn is finite (i.e. it is an element of R rather than ±∞), then it is an accumulation point of {xn}n∈N. 4. Definition A sequence {xn}n∈N ⊂ R is called a Cauchy sequence if for every  > 0, there exists N ∈ N such that |xm − xn| < , for any m,n > N . Definition A infinite series ∞∑ n=1 xn is said to converge if its sequence of partial sums {sN}N∈N converges, where sN := N∑ n=1 xn a) Prove that a sequence of real numbers converges if and only if it is a Cauchy sequence. b) Prove that the harmonic series ∞∑ n=1 1 n does not converge by showing that its sequence of partial sums is not Cauchy. 5. a) Prove that if |x| < 1, then lim xn = 0. (Hint: How do the two numbers lim xn and x · (lim xn) compare? If you use this approach, you will have to prove convergence separately.) b) For each n ∈ N, define a function fn : [−1, 1] −→ R by fn(x) = xn. Compute lim sup fn(x) and lim inf fn(x) for each x ∈ [−1, 1]. M365C (Fall 2006) Problem Set 2 Due: Friday September 15, 2006 2 Sample solutions 1. Let  > 0 be given. Since xn −→ 0, there exists N ∈ N such that |xn| < 2, for each n > N . Hence, √ xn < , for each n > N . Since  > 0 is arbitrary, we conclude that the sequence {√xn} converges and lim √ xn = 0. 2. Let B > 0. Since the sequence {xn}n∈N is unbounded, there exists M ∈ N such that xM > B. Since the sequence is also nondecreasing, we in fact have B < xM ≤ xM+1 ≤ xM+2 ≤ · · · . This shows that xn > B for all n > M . Since B > 0 is arbitrary, we see that indeed lim xn = +∞. 3. Let a := lim supxn ∈ R. Recall that a = lim supxn = lim N→∞ UN = lim N→∞ sup{xN+1, xN+2, . . .}. By the convergence UN → a, there exists some N1 ∈ N such that a− 1 < UN1 := sup{x1+N1 , x2+N1 , . . .} < a + 1. By property of the supremum, there exists some n1 > N1 such that a−1 ≤ xn1 ≤ UN1 < a+1. By the convergence UN → a again, there exists come N2 > n1 such that a− 1 2 < UN2 < a + 1 2 . By property of the supremum again, there exists n2 > N2 (hence n2 > n1) such that a − 1 2 ≤ xn2 ≤ UN2 < a + 1 2 . Continue inductively to obtain a subsequence {xnk}k∈N which satisfies: a− 1 k ≤ xnk ≤ a + 1 k , for each k ∈ N. Obviously, the subsequence xnk converges to a. Hence, a is indeed an accumulation of {xn}, as desired. 4. a) Suppose {xn} converges to x ∈ R. Let  > 0 be given. The convergence xn → x implies there exists N ∈ N such that |xn − x| <  2 , for each n > N . Hence, for each m,n > N , we have |xm − xn| = |xm − x + x− xn| ≤ |xm − x|+ |x− xn| <  2 +  2 < . This proves that convergent sequences of real numbers are Cauchy sequences. Conversely, suppose {xn} is a Cauchy sequence. Then, there exists N ∈ N such that |xm − xn| < 1, for any m,n > N . In particular, xN+1 − 1 < xn < xN+1 + 1, for each n > N . Hence, we have min{xN+1 − 1, x1, . . . , xN+1} ≤ xn ≤ max{xN+1 − 1, x1, . . . , xN+1} for each n ∈ N. This shows that the sequence {xn} is bounded. By the Bolzano-Weierstrass Theorem for R, {xn} has a convergent subsequence {xnk}. Let x := limk→∞ xnk ∈ R. We now prove that the whole sequence {xn} converges to x. So, let  > 0 be given. By the convergence xnk → x, there exists K ∈ N such that |xnk−x| <  2 , for each k ≥ K. By the Cauchy hypothesis on {xn}, there exists N ∈ N such that |xm − xn| <  2 , for any m,n ≥ N . Now, fix any k > max{K, N}. Then, nk ≥ k > max{N,K}. Consequently, for any n > N , we have |xn − x| ≤ |xn − xnk |+ |xnk − x| <  2 +  2 = . Since  is arbitrary, we see that indeed xn → x.