Questions and Answers #17 Series

Questions and Answers Sheet 17 Series Question #1 How can you derive that ∑∞ 𝑛=1 𝑛 3 = 3𝑛 ? 4 Answer: Let 𝑆𝑛 = ∑∞ 𝑛=1 ∞ 𝑛 (1) 3𝑛 ∞ ∞ 𝑛=1 𝑛=1 1 𝑛 𝑛+1 1 𝑆 = ∑ 𝑛+1 = ∑ 𝑛+1 − ∑ 𝑛+1 3 𝑛 3 3 3 𝑛=1 (1)-(2) 2 𝑆𝑛 = ∑∞ 𝑛=1 3 𝑛 = ∑∞ 𝑛=1 𝑛 𝑛+1 1 + ∑∞ 𝑛=1 3𝑛+1 3𝑛+1 𝑛 1 ∞ ∑∞ 𝑛=2 3𝑛 + ∑𝑛=1 3𝑛+1 1 − ∑∞ 𝑛=1 3𝑛 − 3𝑛 1 = + ∑∞ 𝑛=1 3𝑛+1 3 1 1 ∞ = ∑𝑛=1 𝑛 = 3 2 3 Thus, 𝑆𝑛 = 4 Question #2 Show that ∑∞ 𝑛=1 𝑧2 𝑛 𝑛+1 1−𝑧 2 is algebraic. Answer: There is in fact a nice closed form for the partial-sums that you can find just by calculating a bit: 𝑓(𝑧) = = = = = 𝑧2 𝑧4 + 𝑧8 + 1−𝑧 4 1−𝑧 8 𝑧 2 (1+𝑧 4 )+𝑧 4 𝑧8 1−𝑧 8 𝑧 2 +𝑧 4 +𝑧 6 1−𝑧 16 + +⋯ 1−𝑧 16 𝑧8 + +⋯ +⋯ 1−𝑧 8 1−𝑧 16 (𝑧 2 +𝑧 4 +𝑧 6 )(1+𝑧 8 )+𝑧 8 +⋯ 1−𝑧 16 𝑧 2 +𝑧 4 +𝑧 6 +𝑧 8 +𝑧 10 +𝑧 12 +𝑧 14 1−𝑧 16 +⋯ So first few partial sums are given by 𝑓1 (𝑧) = 𝑓2 (𝑧) = 𝑧2 = 𝑧 2 −𝑧 4 1 1−𝑧 4 1−𝑧 2 1−𝑧 4 𝑧 2 +𝑧 4 +𝑧 6 𝑧 2 −𝑧 8 = 1−𝑧 8 1−𝑧 2 𝑧 2 −𝑧 16 1 𝑓3 (𝑧) = ⋯ = 1−𝑧 2 ⋅ ⋅ 1 1−𝑧 8 1−𝑧 16 It should then be easy to show by induction that 𝑓𝑛 (𝑧) = 𝑧2 1−𝑧 2 ⋅ 1−𝑧 2 𝑛−1 1−𝑧 2 𝑛 The second factor is 1 + 𝑂(𝑧 2 𝑧2 𝑓(𝑧) = lim 𝑓𝑛 (𝑧) = 𝑛→∞ 1 − 𝑧2 𝑛−1 ) → 1, as 𝑛 → ∞, thus Question #3 Prove 3𝜋 8 1 𝑘 𝐹2𝑘+1 2𝑘+1 𝑞 5 2𝑘+1 = ∑𝑘≥0 (− ) 2+2√2 where 𝑞 = 17+8√2 1+√ , and 𝐹𝑛 are the Fibonacci numbers. 5 Answer: Let 𝑓(𝑥) = ∑𝑘≥0 𝐹𝑘+1 𝑥 𝑘 so that (1 − 𝑥 − 𝑥 2 )𝑓(𝑥) = 𝐹1 + (−𝐹1 + 𝐹2 )𝑥 or 1 𝑓(𝑥) = 2 1−𝑥−𝑥 Integrating the above we get 𝑥 𝑑𝑡 𝐹 ∑𝑘≥0 𝑘+1 𝑥 𝑘+1 𝐹(𝑥) = ∫0 2 = 1−𝑡−𝑡 𝑘+1 The integrand on left side can be written 1 𝐴 𝐵 = + (1−𝑎𝑡)(1−𝑏𝑡) 1−𝑎𝑡 1−𝑏𝑡 where a, b are roots of 𝑧 2 − 𝑧 − 1 = 0 and 𝐴= 𝑎 𝑎−𝑏 𝑏 ,𝐵 = − 𝑎−𝑏 Letting 𝑎= 1+√5 2 ,𝑏 = 1−√5 2 we get 𝑎 − 𝑏 = √5 and hence we have 𝐹(𝑥) = 1 √5 1−𝑏𝑥 log 1−𝑎𝑥 and 𝐹 𝐺(𝑥) = 𝐹(𝑥) − 𝐹(−𝑥) = 2 ∑𝑘≥0 2𝑘+1 𝑥 2𝑘+1 2𝑘+1 Replacing x by ix we get ∑𝑘≥0(−1)𝑘 𝐹2𝑘+1 2𝑘+1 𝑥 2𝑘+1 = −𝑖 𝐺(𝑖𝑥) 2 The sum in question is −𝑖√5 𝐺(𝑖𝑞/√5) 2 Now 𝐺(𝑥) = 1 √5 1+√5𝑥+𝑥 2 log 1−√5𝑥+𝑥 2 and hence 𝐺(𝑖𝑞/√5) = 1 √5 log 5+5𝑖𝑞−𝑞 2 5−5𝑖𝑞−𝑞 2 = 2𝑖 √5 arctan The desired sum is thus equal to arctan 5𝑞 5−𝑞 2 that 5𝑞/(5 − 𝑞 2 ) = 1 + √2 Question #4 Evaluate: ∑∞ 𝑛=1 arctan ( 1 8𝑛2 ) 5𝑞 5−𝑞 2 Answer: We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is − arctan(1/8𝑛2 ) ∑∞ 𝑛=1 ( 1 8𝑛2 ) = −arg ∏∞ 𝑛=1 (1 − 𝑖 8𝑛2 ) 2 (√𝑖/8) = −arg ∏∞ 𝑛=1 (1 − = −arg ) 𝑛2 sin(𝜋√𝑖/8) 𝜋√𝑖/8 = −arg ( (1−𝑖)√2 cosh(𝜋/4) = − arctan ( + (1+𝑖)√2 sinh(𝜋/4) 𝜋 sinh(𝜋/4)−cosh(𝜋/4) sinh(𝜋/4)+cosh(𝜋/4) 𝜋/2 =arccot (𝑒 ) 𝜋 ) ) Question #5 How can I prove this result? ∑∞ 𝑘=0 2𝑘 𝑘 22 +1 =1 Answer: Make it into a telescopic sum as follows ∑𝑛𝑘=0 𝑘 2𝑘 𝑘 22 +1 = ∑𝑛𝑘=0 = ∑𝑛𝑘=0 2𝑘 (2 (22 2𝑘 𝑘+1 k =0 (22 − 1 =1− 2k k +1 2𝑛+1 (22 𝑛+1 𝑘+1 −1) −1) 2 (2 + 1) = (22 −1) n k 2𝑘 (22 −1) −1) − 2k +1 k +1 (22 − 1) →1 Question #6 Evaluate: lim 𝑛→∞ ∑𝑛 𝑘=1 1 𝑛+√𝑘2 +𝑛 Answer: 1 lim ∑𝑛𝑘=1 𝑛→∞ 𝑛 = lim = 1 1+√(𝑘/𝑛)2 +1/𝑛 1 𝑛 1 ∑ 𝑛 𝑘=1 1+(𝑘/𝑛) = lim lim 1 𝑛→∞ 𝑚→∞ 𝑛 ∑𝑛𝑘=1 1 1+√(𝑘/𝑛)2 +1/𝑚 𝑛→∞ 1 𝑑𝑥 ∫0 1+𝑥 = log 2 The relevant theorem is if 𝑥𝑛𝑚 → 𝑦𝑛 unformly as 𝑚 → ∞ and 𝑥𝑛𝑚 converges as 𝑛 → ∞, then the double limit exists and lim 𝑥𝑛𝑛 = lim lim 𝑥𝑛𝑚 = lim lim 𝑥𝑛𝑚 𝑛→∞ 𝑛→∞ 𝑚→∞ 𝑚→∞ 𝑛→∞ To show uniform convergence in this case, note that | 1 n 1 1 n 1 −  |  2 n k =1 1 + (k / n) + 1/ m n k =1 1 + k / n  1 n |1 + k / n − (1 + (k / n) 2 + 1/ m  n k =1 |1 + k / n ||1 + (k / n) 2 + 1/ m |  1 n ( (k / n) 2 − 1/ m − k / n)  n k =1 1 1/𝑚 𝑛 1 √(𝑘/𝑛)2 +1/𝑚+𝑘/𝑛 1/𝑚 𝑛 1 1/√𝑚 = ∑𝑛𝑘=1 ≤ ∑𝑛𝑘=1 = √𝑚 1 showing that lim 𝑚→∞ 𝑛 √𝑘 = 1𝑛 1 1+√(𝑘/𝑛)2 +1/𝑚 1 1 𝑛 1+𝑘/𝑛 = ∑𝑛𝑘=1 uniformly for all n. Question #7 Find: ∑∞ 𝑘=1 6𝑘 (3𝑘+1 −2𝑘+1 )(3𝑘 −2𝑘 ) Answer: That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form ∑𝑘≥1 ( 𝐴𝑘 3𝑘 −2𝑘 − 𝐴𝑘+1 3𝑘+1 −2𝑘+1 ) where 𝐴𝑘 very likely depends on k. Note that if this works, the sum of the series will be 𝐴1 31 −21 = 𝐴1 Now Ak Ak +1 3k +1 Ak − 3k Ak +1 − 2k +1 Ak + 2k Ak +1 − = 3k − 2k 3k +1 − 2k +1 (3k − 2k )(3k +1 − 2k +1 so you want to choose 𝐴𝑘 and 𝐴𝑘+1 so that 3𝑘+1 𝐴𝑘 − 3𝑘 𝐴𝑘+1 − 2𝑘+1 𝐴𝑘 + 2𝑘 𝐴𝑘+1 = 6𝑘 The obvious things to try are 𝐴𝑘 = 2𝑘 , which makes the last two terms cancel out to leave 3𝑘+1 2𝑘 − 3𝑘 2𝑘+1 = 6𝑘 (3 − 2) = 6𝑘 , and 𝐴𝑘 = 3𝑘 , which makes the first two terms cancel out and leaves 6𝑘 (3 − 2) = 6𝑘 ; both work. However, summing ∑𝑘≥1 ( 𝐴𝑘 3𝑘 −2𝑘 − 𝐴𝑘+1 3𝑘+1 −2𝑘+1 ) (1) to 𝐴1 31 −21 = 𝐴1 is valid only if 𝐴 lim 𝑘 𝑘 𝑘 = 0 𝑘→∞ 3 −2 since the n-th partial sum of (1) is 𝐴 𝐴1 − 𝑛+1𝑛+1𝑛+1 3 −2 Checking the two possibilities, we see that lim 2𝑘 𝑘→∞ 3𝑘 −2𝑘 =0 , but lim 3𝑘 𝑘→∞ 3𝑘 −2𝑘 =1 so we must choose 𝐴𝑘 = 2𝑘 , and the sum of the series is indeed 𝐴1 = 2. Question #8 Consider, for all 𝑛 ∈ 𝑁 : 1 𝑢𝑛 = ∫0 (1 − (1 − 𝑡 𝑛 )1/𝑛 )𝑑𝑡 Does the series ∑𝑛≥1 𝑢𝑛 converges ? Answer: We have 1 (1 − 𝑡 𝑛 )1/𝑛 = 𝑒 𝑛 log(1−𝑡 1 ≥ 1 + log(1 − 𝑡 𝑛 ) 𝑛) 𝑛 from which 1 1 1 ∫0 (1 − 𝑡 𝑛 )1/𝑛 𝑑𝑡 ≥ 1 + ∫0 log(1 − 𝑡 𝑛 ) 𝑑𝑡 𝑛 =1+ ≥ 1+ =1− 1 log(1−𝑡) 1/𝑛 𝑡 ∫ 𝑛2 0 𝑡 1 1 log(1−𝑡) 𝑑𝑡 ∫ 𝑛2 0 𝑡 2 𝜋 1 𝑑𝑡 6𝑛2 1 Hence, 0 ≤ ∫0 (1 − (1 − 𝑡 𝑛 )1/𝑛 )𝑑𝑡 ≤ 𝜋2 6𝑛2 By the comparison test, the series converges. Question #9 Does the sum ∑∞ 𝑛=1 √𝑛! (3+√1)(3+√2)…(3+√𝑛) converge or diverge? Answer: Let 𝑎𝑛 ≔ √𝑛! (3 + √1)(3 + √2) … (3 + √𝑛) = 1 (1 + 3/√1)(1 + 3√2) … (1 + 3/√𝑛) Then log(𝑎𝑛 ) = − ∑𝑛𝑘=1 log (1 + 3 √𝑘 ) In the sum, we have log(1 + 3/√𝑘) ∼ log(𝑎𝑛 ) ∼ − ∑𝑛𝑘=1 3 √𝑘 3 √𝑘 as 𝑘 → ∞, so ∼ −6√𝑛 as 𝑛 → ∞. Now, it follows that log(𝑎𝑛 ) log 𝑛 n sufficiently large, 𝑎𝑛 < → −∞ 𝑎𝑠 𝑛 → ∞, so for n sufficiently large, we have 1 𝑛2 log(𝑎𝑛 ) log 𝑛 < −2. Therefore, for , and by the comparison test we can conclude that ∑∞ 𝑛=1 𝑎𝑛 converges. Question #10 I want to show that ∑𝑛−1 𝑘=1 1 2𝑘𝜋 ) 𝑛 1−cos( = 𝑛2 −1 6 Answer: 1 If 𝑧 = 𝑒 𝑖𝑡 , 𝑧 ≠ 1, then cos 𝑡 = (𝑧 + 𝑧 −1 ) hence 1/(1 − cos 𝑡) = −2𝑧/(1 − 𝑧)2 . The sum 𝑠𝑛 to be 2 computed is −2𝑧 𝑠𝑛 = ∑∗𝑧 2 (1−𝑧) where ∑∗𝑧 means that the sum runs over every n-th root z of 1 different from 1. For every |x|<1, consider 𝑧𝑥 𝑡𝑛 (𝑥) = ∑𝑧 2 (1−𝑧𝑥) where ∑𝑧 means that the sum runs over every n-th root z of 1. Expanding each ratio 𝑧𝑥/(1 − 𝑧𝑥)2 as a power series in (zx), one gets 𝑡𝑛 (𝑥) = ∑𝑘≥1 𝑘𝑢𝑘𝑛 𝑥 𝑘 , 𝑢𝑘𝑛 = ∑𝑧 𝑧 𝑘 Now, 𝑢𝑘𝑛 = 0 for every k except when k is a multiple of n, in which case 𝑢𝑘𝑛 = 𝑛. Thus 𝑡𝑛 (𝑥) = 𝑛2 ∑𝑘≥1 𝑘𝑥 𝑛𝑘 𝑛2 𝑥 𝑛 = (1−𝑥 𝑛)2 This allows to compute 𝑠𝑛 since 𝑥 𝑠𝑛 = 2 ⋅ lim ((1−𝑥)2 − 𝑡𝑛 (𝑥)) 𝑥→1 Expanding both terms in the limit in powers of 1-x when 𝑥 → 1 yields finally 𝑠𝑛 = 2 ⋅ 𝑛2 −1 12 = 𝑛2 −1 6