Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 2.6.3, 2.7.4, 2.7.5, 2.7.2, 2.7.3 Problem A. Compute. Hint: Use the identity n 2 n 2 = 2 n, n + and examine the partial sums. Problem B. Determine the convergence or divergence of the series 2 n n!. n= Recall that n! indicates the product n! = n n n 2... 2. Problem C. Suppose the series a n is a convergent series, and let t m indicate n= t m = a n. n=m That is, t m is a tail of the original series. Show that the series t m converges. Hint: Cauchy s Criterion for series. 2 Show that the sequence {t m } converges to 0. Solution 2.6.3a. A pseudo-cauchy sequence must have the distance between large enough successive terms become arbitrarily small,
2 whereas a Cauchy sequence does not insist that these terms be successive. Solution 2.6.3b. Let s k indicate the kth partial sum of the harmonic series. Then the sequence of partial sums {s k } is pseudo-cauchy: For each ɛ > 0, there is an N N so that for all n N we have /n < ɛ. In that case, the difference s n+ s n = /n is less than ɛ for all n N. On the other hand, the harmonic series diverges, so the sequence {s k } does not converge, and is not Cauchy by the Cauchy Criterion. Solution 2.7.4. Let x n = y n = /n, so that each of the series x n and y n are divergent. On the other hand, x n y n = /n 2, and we ve seen that /n 2 is convergent for example, by the p-test. Solution 2.7.5 a. Let s n and t n indicate the partial sums s n = a +... + a n and t n = a 2 +... + a 2 n. By assumption, {s n } converges, and is thus Cauchy. In order to see that a 2 n converges, by the Cauchy Criterion again it suffices to prove that {t n } is Cauchy. Thus, given ɛ > 0, we d like to make t n t m less than ɛ. For n > m, we have t n t m = a 2 m+ +... + a 2 n. We claim that, for any non-negative numbers x,..., x n, we have x 2 +... + x 2 n x +... + x n 2. We prove the latter statement by induction: For n =, the statement is obvious. Invoking the inductive hypothesis, we have x 2 +... + x 2 n+ x 2 +... + 2x +... + x n x n+ + x 2 n+ x +... + x n 2 + 2x +... + x n x n+ + x 2 n+ = x +... + x n+ 2. We may now apply this fact in our setting, noting that a,..., a n are all non-negative. For any ɛ > 0, there exists N N so that for all m, n N we have s m s n < ɛ. For any m, n N, we have t n t m = a m+ 2 +... + a n 2 a m+ +... + a n 2 = s n s m 2 < ɛ 2 = ɛ, so that {t n } is Cauchy, as desired.
Solution 2.7.5b. No, we may not conclude that an converges, even though a n > 0. For example, the series /n 2 converges, while the harmonic series does not. Solution 2.7.2. Since x j = s j s j, we have: x j y j = s j s j y j = s j y j s j y j By replacing j by a new dummy variable j = j, so that j goes from m to n as j goes from m + to n, we can rewrite the last expression as n s j y j s j y j +. j =m Finally replacing j with j, we see: n x j y j = s j y j s j y j+ as desired. = s j y j = s n y n+ s m y m+ + j=m s m y m+ s n y n+ + s j y j y j+, s j y j+ Solution 2.7.3a. Using the previous exercise and the triangle inequality, we see that x j y j = s ny n+ s m y m+ + s j y j y j+ s n y n+ + s m y m+ + s j y j y j+. Since y j 0 for all j, and s n M for all n, the last expression above is less than My n+ + My m+ + My j y j+.. 3
4 The last sum above is telescoping, and after canceling we find x j y j My n+ + y m+ + y m+ y n+ = 2My m+, as desired. Solution 2.7.3b. Let t k be the kth partial sum of x n y n. Note that, for n > m, we have t n t m = x j y j. For each ɛ > 0, there is an N so that, for all n N we have y n = y n < ɛ/2m, since {y n } converges to 0. For any n > m N, we have t n t m = x j y j 2My m+ < 2M ɛ 2M = ɛ, so that {t n } is a Cauchy sequence. By Cauchy s Criterion, this sequence converges, and the original series converges. Solution 2.7.3c. Let x n = n+, so that the partial sums of xn are all equal to or. Thus the partial sums are all bounded. Letting y n = a n, the assumptions in the Alternating Series Test say that {y n } is monotone decreasing, non-negative, and converges to 0. By Dirichlet s Test, x n y n converges. Solution A. Let s N indicate the partial sum N n 2.
5 Using the hint, we find N s N = 2 n n + N N n = 2 = 2 = 2 n + + 2 + N N + + N + 2 N N +. n=4 N 2 n n + Using the algebraic limit properties of limits and the fact that lim /n = 0, it is straightforward to see that lim s N = + = N 2 2 2 3 2 = 3 4. Solution B. We compute the ratio of successive terms: 2 n+ n +! n! 2 = 2 n n. The latter has limit 0, so the the ratio test proven as exercise 2.7.9 on homework 3 applies: As 0 <, the series converges. Solution C. Let s indicate the partial sums of the series a n by and those of t m by s k = v m,k = k a n, n= m+k n=m Since the series a n converges, the sequence {s k } converges, and it is thus Cauchy by the Cauchy criterion. We conclude that, for any ɛ > 0, there is an N so that, for all k > l N we have k a n = s k s l < ɛ. n=l+ a n.
Note that, for each m and k, we have v m,k = m+k n=m Thus, for k > l, we have a n = m+k n= a n m n= = s m+k s m. v m,k v m,l = s m+k s m s m+l s m = s m+k s m+l. Suppose now that k > l N +, so that k + m > l + m l N. By the choice of N above, v m,k v m,l = s m+k s m+l < ɛ, so that {v m,k } is a Cauchy sequence. By Cauchy s Criterion, the series t m converges. Solution C2. We restate the Cauchy condition for {s k }: For each ɛ > 0, there is an N so that for all k, l N, s k s l < ɛ. For m N +, we have m + k > m N, so that s m+k s m < ɛ. On the other hand, this implies that v m,k = s m+k s m < ɛ. In particular, the convergent sequence {v m,k } is bounded in absolute value by ɛ. By the Order Limit Theorem p. 48, we conclude that the limit, which is by definition the value of t m, is bounded in absolute value by ɛ. We have shown that, for any ɛ > 0, there is an N so that for all m N we have t m < ɛ. Thus {t m } converges to 0.