Solutions for Homework Assignment 5. Problem. Assume that a n is an absolute convergent series. Prove that absolutely convergent for all x in the closed interval [, ]. a n x n is Solution. Assume that a n isabsoluteconvergent. Thus, a n isaconvergent infinite series. Its terms are obviously nonnegative. Suppose that x [. ]. We then have x. It follows that = a n x n a n n = a n for all n. Since and a n are nonnegative, a n for all n, and a n is convergent, we can use Comparison Theorem I to conclude that Therefore, the infinite series is convergent. a n x n is absolutely convergent for all x [, ]. Problem 2. For each of the following power series, determine precisely the set of values of x for which the power series converges. A general remark: Suppose that {b n } is a sequence and that b n =. It follows that if F(x) is a continuous function, then F(b n) = F(). (We stated this theorem in class one day.) Take F(x) = x, which is indeed a continuous function. We obtained the following useful result: If b n =, then b n =. We will use this result several times in the solutions below. It will be used in the contrapositive form.
(a) x n n 2. Solution. Theseriesconvergesforx [,]. Thisfollowsfromproblembecause n 2 is known to be convergent. Since the terms are nonnegative, that series is also absolutely x n convergent. Therefore, by problem, is absolutely convergent for all x [, ]. n2 By theorem 3 on the handout about infinite series, it follows that all x [. ]. x n is convergent for n2 If x [, ], then x >. Thus, / x <. Using a fact mentioned in class on Monday, it follows that n2 (/ x ) n =. This fact implies that < n 2 (/ x ) n < for x n all sufficiently large n. It follows that > for all sufficiently large n. Therefore, it n 2 is not true that xn =. According to the general remark stated at the beginning n 2 x n of problem 2, it is also not true =. Hence, by theorem on the infinite series n2 x n handout, the infinite series diverges if x [, ]. n2 (b) n! n nxn Solution. We will show that this infinite series converges if and only if x ( e, e). It obviously converges if x =. And so we can concentrate in the rest of this solution on the case where x. The terms in the infinite series in question are then nonzero. Let a n = n! n n. Then / ) ) (n+) n+ = ( (n+)! n! n n ( / ( = + ) ) n x. n x n+ x 2 = ( (n+) /(n+) (n+)n n n x
It follows that = e x If x < e, then < and therefore the series a a n x n n x n converges. Therefore, the infinite series being considered in this question is absolutely convergent when x < e. It isthereforeconvergent when x < e. (Weareusing theorem3ontheinfinite series handout.) Equivalently, the infinite series in question converges when x ( e, e). Now assume that x e. Then n! n nxn n! = n nen ( n n n!e n ). We recall a result proved in problem 5 of Homework Assignment 2. Let s n = nn n!en. It was proved that the sequence {s n } is strictly decreasing. In particular, we have s n s = /e for all n. Noting that s n > for all n, it follows that s n e for all n. Therefore, if x e, then we have ( ) n! n n n nxn = s n!e n n e for all n. It follows that the sequence { n! x n } doesn t converge to when x e. n n Therefore, the sequence { n! x n } also doesn t converge to when x e. By theorem n n on the infinite series handout, it follows that the infinite series in this question diverges if x e. n! We have proved that the infinite series converges when x ( e, e), and n nxn diverges for all other values of x. (c) n3 x n. Solution. As in part (b), the infinite series obviously converghes when x = and so we may assume in the rest of this solution that x. Let a n = n 3. Then = 3 ( ) 3 n+ x n
and = 3 x = x It follows from theorem 5 on the infinite series handout that n 3 x n converges absolutely, and therefore converges, when x <. (Of course, this is true even when x =.) If x, then n 3 x n = n 3 x n n 3 n = n 3 for all n. Therefore, it is not true that n 3 x n =. Thus, by theorem on the infinite series handout, n 3 x n cannot be convergent when x. The given infinite series converges if and only if x <. (d) n!x n. Solution. We will show that this infinite series converges if and only if x =. If x =, all the terms in the series are equal to. The series obviously converges when x =. Now assume that x. Let a n = n!. Then = (n+)! n! x n+ x n = (n+) x and hence = +. since x >. Therefore, there exists a positive integer N such that n N = a n+x n+ > and hence > when n N. It follows that a m x m > a N x N for all m > N. Since a N x N >, the sequence { a m x m } cannot converge to. Therefore, it is also true that the sequence {a m x m } cannot converge to. 4
According to theorem on the infinite series handout, it follows that the infinite series in this question cannot converge if x. In summary, the infinite series n!x n converges only when x =. Section 2., Exercise (a). This question concerns the sequence {f n (x)} where f n (x) = (sin x) n for all n. We consider the interval [,π]. Solution. Note that sin x < for all x [,π] except for the point x = π/2. Thus, it follows that f n(x) = if x [,π] and x π/2. If x = π/2, then f n (x) = n = for all n. Therefore, the sequence {f n (x)} converges to the function f(x) on the interval [,π], where if x [,π], x π/2 f(x) = if x = π/2 Wewill show that thesequence {f n (x)} doesnot converge uniformlyontheinterval [,π]. Assume to the contrary that the sequence {f n (x)} does converge uniformly on the interval [,π]. The limit function is the function f(x) defined above. Let ǫ =. Then, since we are 2 assuming that the convergence is uniform, there exists a number N such that () n N = f n (x) f(x) < 2 for all x [,π]. Suppose that n is an integer satisfying n N. Note that n 2 > and hence < / n 2 <. Let x n = arcsin(/ n 2). Then < x n < π/2. We have f n (x n ) = ( sin(x n ) )n = ( / n 2 )n = /2 and f(x n ) = Therefore, we have f n (x n ) f(x n ) = (/2) = 2. However, since n N and x n [, π], statement () implies that f n (x n ) f(x n ) <. 2 It follows that <, which is not true. And so we have a contradiction. Therefore, the 2 2 sequence {f n (x)} cannot converge uniformly on the interval [,π]. 5
Section 2., Exercise (d). f n (x) = nxe nx for all x. This question concerns the sequence {f n (x)}, where Solution. Notice that f n () = for all n and hence {f n ()} converges to as n. If x >, then n (e n ) =, x a fact which was mentioned in class on Monday. It follows that f n(x) = x n (e n ) = x = x forallx >. Therefore, thesequence {f n (x)} converges tothefunctionf(x), where f(x) = for all x. It will be useful to consider the function g(t) defined by g(t) = te t = t/e t for all t. Note that the derivative is given by g (t) = et te t = ( t)e t. e 2t It follows that g (t) < for t >. Hence g(t) is a decreasing function for t >. Thus, if u v >, then g(u) g(v). Suppose that δ >. Pick a positive integer N so that N δ >. Then, if n N and x δ, we have nx nδ N δ >. Therefore, since g(t) is decreasing for t >, it follows that g(nx) g(nδ) for all n N. Notice that g(nx) = nxe nx = f n (x) and g(nδ) = f n (δ). Therefore, if n N, we have < f n (x) f n (δ) for all x δ. Since the sequence {f n (x)} converges to f(x) for all x, it follows that {f n (δ)} converges to f(δ). Note that f(δ) =. Suppose that ǫ >. We know that f n (δ) =. Therefore, there exists an N 2 > such that n N 2 = f n (δ) < ǫ Since f n (x) for all x, we have f n (δ) = f n (δ) and f n (x) f(x) = f n (x) = f(x) for all x. We will use these facts below. Let N = max{n, N 2 }. Then n N = n N and n N 2 = < f n (x) f n (δ) < ǫ for all x δ. 6
Thus, we have n N = f n (x) f(x) < ǫ for all x δ. This means that the sequence {f n (x)} converges uniformly to f(x) on the interval [δ, ). Now we consider the interval [, δ]. We will show that the convergence is not uniform. Assume to the contrary that {f n (x)} converges uniformly to f(x) on the interval [, δ]. Let x n = /n. Notice that f n (x n ) = n(/n)e n(/n) = /e. Let ǫ = /2e. Then ǫ >. Since we are assuming that {f n (x)} converges uniformly to f(x) on the interval [,δ], there must exist a positive number N with the following property: n N = f n (x) f(x) < /2e for all x [, δ]. Choose an integer n > Max{/δ,N}. Then n > /δ and we have < /n < δ. Hence x n [, δ]. Therefore, since n N, it follows that Hence it also follows that f n (x n ) f(x n ) < /2e /e = /e < /2e. Of course, this inequality is not true. We have a contradiction. Therefore, the sequence {f n (x)} cannot converge uniformly to f(x) on the interval [, δ]. Section 2., Exercise (f). This question concerns the sequence of functions defined by f n (x) = (x/n)e x/n for all n. We fix a positive number A. Solution. Notice that e x/n for all x. Thus, we have f n (x) x/n for all x. Since, for any x, we have x/n =, it follows that f n(x) = 7
Therefore, the sequence {f n (x)} converges to the function f(x) defined by f(x) = for all x. Furthermore, if x A, then we have Suppose that ǫ >. Notice that f n (x) A/n. A/n < ǫ A/ǫ < n Choose N = A/ǫ+. Then, for all x [, A], we have n N = n > A/ǫ = A/n < ǫ = f n (x) < ǫ for all x [, A]. We have used the fact pointed out before that f n (x) A/n for all x [, A]. Now f(x) = for all x. Therefore, for the N chosen above, we have n N = f n (x) f(x) < ǫ for all x [, A]. As a consequence, the sequence {f n (x)} converges uniformly to f(x) on the interval [, A]. This is true for any number A >. In contrast, it is not true that the sequence {f n (x)} converges uniformly to f(x) on the interval [, ). This can be seen by considering the point x n = n for any n. We have f n (x n ) = e. Let ǫ be any positive number such that ǫ < e. Then, for every n (no matter how large), the inequality f n (x) f(x) < ǫ fails to be satisfied for at least one x [, ). Namely, it fails to be satisfied for x = x n = n. Section 2.2, Exercise 3. This problem concerns p >. Solution. First of all, recall that function ζ(x) on the interval (, ) by nx. Let p R and assume that converges for all x >. Thus, we can define a nx ζ(x) = 8 n x.
This is an important function in mathematics and is usually called the Riemann ζ-function. We will use the M-Test. This is Theorem II in section 2.2. It is stated on the handout about uniform convergence. Note that if n, then we have the inequality n x n p for all x p. Therefore, n x n p for all x [p, ) Furthermore, note that the infinite series converges. This is true because p >. Let M n = for all n. We have np n p < n x M n for all x [p, ). The infinite series M n converges. The assumptions in the M-test are satisfied. We can therefore conclude that converges uniformly to ζ(x) on the interval [p, ). nx Section 2.4, Exercise 3. This question concerns the sequence {f n (x)}, where f n (x) = nxe nx2 for all x [,]. Solution. Note that f n () = for all n. Hence f n () =. Furthermore, if x >, then ( ) n f n (x) = x. (e n ) x As mentioned in class on Monday, if x >, then ( ) n (e n ) x =. It follows that f n(x) = when x >. This is also true when x =, as we pointed out above. Thus the sequence {f n (x)} converges to the function f(x) defined by f(x) = for all x. Obviously, we have f(x)dx =. Let g n (x) = 2 e nx2. Then, according to the chain rule, g n(x) = f n (x). Hence g n (x) is an antiderivative for the function f n (x). Therefore, f n (x)dx = g n () g n () = 2 e n ( 2 9 ) ( ) = e n 2
for all n. It follows that ( f n (x)dx = ) e n = /2 2 In contrast, we have f(x)dx =. Hence, f n (x)dx f(x)dx. This is what we intended to show. We can add the following remark. The functions f n (x) are continuous on the interval [,] for all n. We can then use the Integration Theorem (in its contrapositive form) to conclude that the sequence {f n (x)} does not converge uniformly on the interval [, ].