M 3001 ASSIGNMENT 5: SOLUTION Due in class: Wednesday, Feb. 16, 2011 Name M.U.N. Number P∞ P∞ P∞ 1. Suppose n=1 an = ∞. Let n=1 bn be a regrouping of n=1 an formed by inserting paren- P∞ theses. Prove that n=1 bn = ∞. P∞ P∞ Solution. Since n=1 bn is a regrouping of n=1 an, there is an increasing function φ : N → N such that bn = aφ(n). Note that for any M > 0 there is an N ∈ N such that n > N implies Pn Pn Pφ(n) Pn P∞ k=1 ak > M. So k=1 bk = k=1 ak ≥ k=1 ak > M. That is to say, n=1 bn = ∞. P∞ P∞ P∞ 2. If n=1 an converges absolutely and n=1 bn is a rearrangement of n=1 an, prove that P∞ P∞ n=1 |bn| = n=1 |an|. P∞ P∞ P∞ Solution. Since n=1 |an| converges and n=1 |bn| is a rearrangement of n=1 |an|. By the P∞ P∞ theorem on series rearrangement proved in Section 2.3, n=1 |bn| equals n=1 |an|. 3. Give an example showing that a rearrangement of a divergent series may diverge or converge. Solution. Since a series is a rearrangement of itself, a divergent series is its own divergent P∞ rearrangement. Suppose n=1 an is conditionally convergent. Then there is an arrangement P∞ P∞ P∞ n=1 bn = ∞ of n=1 an. In fact, let p1, p2, ... be the nonnegative terms of n=1 an in the P∞ order in which they occur; let −q1, −q2, ... denote the negative terms of n=1 an in the order P∞ P∞ in which they occur; and note that n=1 pn = ∞ = n=1 qn. To construct a rearrangement Pn1 with sum ∞, let n1 be the smallest positive integer such that k=1 pk > 1; let n2 be the smallest positive integer greater than n1 such that n n X1 X2 pk − q1 + pk > 2; k=1 k=n1+1 let n3 be the smallest positive integer greater than n2 such that n n n X1 X2 X3 pk − q1 + pk − q2 + pk > 3. k=1 k=n1+1 k=n2+1 P∞ Continuing (by induction) we obtain the desired rearrangement n=1 bn = ∞. Hence, P∞ P∞ n=1 an, as an arrangement of n=1 bn = ∞, is convergent. P∞ n 4. Prove that the Cauchy product of the alternating harmonic series n=0(−1) /(n + 1) with itself converges. [Hint: Use 1 1 1 1 = + (k + 1)(n − k + 1) n + 2 k + 1 n − k + 1 to obtain ∞ ∞ n X X (−1) 1 1 1 cn = 2 1 + + + ··· + . n=0 n=0 n + 2 2 3 n + 1 Now use the Alternating Series Test to obtain convergence. To get the nth term going to 0, use upper and lower sums for f(x) = 1/x to obtain 1 1 1 ln(n + 2) ≤ 1 + + + ··· + ≤ 1 + ln(n + 1). 2 3 n + 1 Solution. First of all, by induction we can easily check n 1 1 n+1 1 X + = 2 X . k=0 k + 1 n − k + 1 k=1 k Next, we calculate n n n n+1 X n X 1 (−1) 2 X 1 cn = akan−k = (−1) = , k=0 k=0 (k + 1)(n − k + 1) n + 2 k=1 k whence deriving ∞ ∞ n X X (−1) 1 1 1 cn = 2 1 + + + ··· + . n=0 n=0 n + 2 2 3 n + 1 Pn+1 −1 To simplify notation, let tn+1 = k=1 k . Consider f(x) = 1/x for x ∈ [1, n + 2]. Then Z n+2 n+1 Z k+1 −1 X −1 ln(n + 2) = x dx = x dx ≤ tn+1. 1 k=1 k Meanwhile, Z n+1 n Z k+1 n+1 ln(n + 1) = x−1dx = X x−1dx ≥ X k−1. 1 k=1 k k=2 Hence, ln(n + 2) ≤ tn+1 ≤ 1 + ln(n + 1) and so ln(n + 2) t 1 + ln(n + 1) ≤ n+1 ≤ . n + 2 n + 2 n + 2 By an application of L’Hospital’s Rule, the both sides of the last estimates tend to 0. Thus, tn+1 the Squeeze Theorem gives n+2 → 0. tn+1 ∞ −1 Note that ( n+2 )n=1 is monotone decreasing – in fact, tn+2 = tn+1 + (n + 2) implies t t t 1 t n+2 ≤ n+1 ⇔ n+1 + ≤ n+1 ⇔ 1 ≤ t . n + 3 n + 2 n + 3 (n + 2)(n + 3) n + 2 n+1 P∞ So, the Alternating Series Test gives that n=0 cn converges. 5. Prove that the Cauchy product of two absolutely convergent series converges absolutely. P∞ P∞ P∞ Solution. Let n=0 an and n=0 bn converge absolutely, and as usual let n=0 cn denote their P∞ Pn Cauchy product. To prove n=0 |cn| < ∞, let dn = k=0 |ak||bn−k|. By the theorem proved P∞ P∞ P∞ in Section 4, n=0 dn < ∞ since this is the Cauchy product of n=0 |an| and n=0 |bn|. Now P∞ |cn| ≤ dn. So, by the Comparison Test, n=0 |cn| < ∞, as desired. P∞ −n 6. Find the sum of n=1 n3 . P∞ −n Solution. Using the Ratio Test, we see that n=1 n3 is convergent. Moreover, ∞ ∞ ∞ X n3−n = 3−1 X n31−n = 3−1 X(n + 1)3−n = 3/4. n=1 n=1 n=0.