Questions and Answers #9 Series

Questions and Answers Sheet 9 Series Question #1 Proof that: ∑∞ 𝑛=0 1 𝑛! =𝑒 Answer: By definition, 1 𝑛 𝑒 = lim (1 + ) 𝑛 𝑛→∞ Using the binomial theorem, the k-th term of the development is ((𝑛), (𝑘)) 1 𝑛(𝑛−1)(𝑛−2)…(𝑛−𝑘+1) = 𝑛𝑘 and lim ((𝑛), (𝑘)) 𝑛→∞ 1 𝑘!⋅𝑛⋅𝑛…𝑛 1 𝑛𝑘 𝑘! = For example, 1 1000 1 1 0.99 0.997002 0.994010994 (1 + ) = + + + + … 1000 0! 1! 2! 3! 4! ′ Question #2 Prove that: ∑∞ 𝑛=0 (−1)𝑛 (2𝑛+1)2𝑚+1 = (−1)𝑚 𝐸2𝑚𝜋2𝑚+1 4 𝑚+1 (2𝑚)! Answer: Because it wasn't explicitly stated, I probably should mention that 𝐸2𝑚 are the Euler numbers. It can be shown using contour integration. 𝜋 csc 𝜋𝑧 Let 𝑓(𝑧) = (2𝑧+1)2𝑚+1 1 1 2 2 Then integrating around a square with vertices at ± (𝑁 + ) ± 𝑖 (𝑁 + ) (where N is a positive integer greater than 1) and letting N go to infinity, 1 ∑∞ 𝑛=−∞ 𝑅𝑒𝑠[𝑓(𝑧), 𝑛] + 𝑅𝑒𝑠 [𝑓(𝑧), − ] 2 1 (−1)𝑛 = 2 ∑∞ 𝑛=0 (2𝑛+1)2𝑚+1 + 𝑅𝑒𝑠 [𝑓(𝑧), − ] = 0 2 But notice that 1 1 𝑅𝑒𝑠 [𝑓(𝑧), − ] = 𝑅𝑒𝑠 [𝑓 (𝑧 − ) , 0] 2 −𝜋 sec 𝜋𝑧 =Res[ (2𝑧)2𝑚+1 2 ,0 And using the Taylor expansion of sec(𝑧) at the origin, that is, sec 𝑧 = ∑∞ 𝑘=0 (−1)𝑘 𝐸2𝑘 (2𝑘)! 𝑧 2𝑘 we have −𝜋 sec 𝜋𝑧 (2𝑧)2𝑚+1 = −𝜋 22𝑚+1 ∑∞ 𝑘=0 (−1)𝑘 𝐸2𝑘 (2𝑘)! 𝜋 2𝑘 𝑧 2𝑘−2𝑚−1 which implies −𝜋 sec 𝜋𝑧 𝑅𝑒𝑠 [(2𝑧)2𝑚+1 , 0] = −𝜋(−1)𝑚𝐸2𝑚 22𝑚+1 (2𝑚)! 𝜋 2𝑚 Therefore, (−1)𝑛 2 ∑∞ 𝑛=0 (2𝑛+1)2𝑚+1 = (−1)𝑚 𝐸2𝑚 22𝑚+1 (2𝑚)! and the result follows. 𝜋 2𝑚+1 Question #3 Prove (−1)𝑘 csc(𝑥) = ∑∞ 𝑘=−∞ 𝑥+𝑘𝜋 Answer: We can use also this well known summation formula (a consequence of the residue theorem): ∑𝑘∈𝑍(−1)𝑘 𝑓(𝑘) = − ∑[residues of 𝜋 csc(𝜋𝑧) 𝑓(𝑧) 𝑎𝑡 𝑓(𝑧)′ 𝑠 poles] 1 so if we take 𝑓(𝑧) = ∑𝑘∈𝑧 (−1)𝑘 𝑥+𝑘𝜋 we get 𝑥+𝑧𝜋 = −𝑅𝑒𝑠𝑧=−𝑥/𝜋 ( 𝜋 csc(𝜋𝑧) 𝑥+𝑧𝜋 ) = csc(𝑥) as wanted. Question #4 How to sum this series for 𝜋/2 directly? 𝜋 2 𝑘! = ∑∞ 𝑘=0 (2𝑘+1)! Answer: First, (2𝑘 + 1)! = (2𝑘 + 1)(2𝑘 − 1) (2𝑘 + 1)! (2𝑘 + 1)! = = 2𝑘 𝑘! (2𝑘)(2(𝑘 − 1)) … 2(1) So your sum can be rewritten as ∞ ∞ ∑ 𝑘=0 𝑘! 𝑘! 2𝑘 2𝑘 =∑ (2𝑘 + 1)! (2𝑘 + 1)((2𝑘), (𝑘)) 𝑘=0 Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For 4𝑘 −1 example, this paper by Sprugnoli gives the ordinary generating function of 𝑎𝑘 = (2𝑘+1) ((2𝑘), (𝑘)) 1 𝑡 𝑡 1−𝑡 𝐴(𝑡) = √ arctan √ 𝑡 1−𝑡 Subbing in 𝑡 = 1/2 says that ∞ ∑ 𝑘=0 2𝑘 (2𝑘 + 1)((2𝑘), (𝑘)) = 2 arctan(1) = 𝜋 2 Question #5 Is it true that for 𝑥 ∈ [0,2𝜋] we have ∑∞ 𝑛=1 cos(𝑛𝑥) 𝑛2 = 𝑥2 4 − 𝜋𝑥 2 + 𝜋2 6 Answer: I show that ∑∞ 𝑛=−∞ ∑∞ 𝑛=1 1 𝑛2 = 𝜋2 sin2 𝑎𝑛 𝑛2 = 𝜋𝑎 when 𝑎 ∈ (0, 𝜋). Now, use the fact that 6 and let S be the sum in question. Then 𝜋2 6 − 𝑆 = ∑∞ 𝑛=1 1−cos 𝑛𝑥 𝑛2 = 2 ∑∞ 𝑛=1 Rewrite the last sum as sin2 𝑛𝑥/2 𝑛2 to be ∞ ∞ sin2 𝑛 𝑥/2 sin2 𝑛 𝑥/2 𝑥 2 2∑ = ∑ − ( ) 𝑛2 𝑛2 2 𝑛=1 𝑥 2 Then 𝜋2 6 𝑛=−∞ 𝑥2 =𝜋 − 4 𝑥 𝑥2 2 4 −𝑆 = 𝜋 − →𝑆= 𝑥2 4 𝑥 𝜋2 2 6 −𝜋 + Question #6 How can we compute the following limit: lim (1 + 1/𝑛2 ) (1 + 2/𝑛2 ) … (1 + 𝑛/𝑛2 ) 𝑛→∞ Answer: We will estimate 𝑘 𝑓(𝑛) = ∑𝑛𝑘=1 log (1 + 𝑛2 ) Recall that log(1 + 𝑥) = 𝑥 + 𝑂(𝑥 2 ). Hence, we get that 𝑛 𝑓(𝑛) = ∑ ( 𝑘=1 𝑘 𝑘2 𝑛+1 + 𝑂 ( )) = + 𝑂(1/𝑛)(1) 𝑛2 𝑛4 2𝑛 where we made use of the fact that ∑𝑛𝑘=1 𝑘 = 𝑛(𝑛+1) 2 and ∑𝑛𝑘=1 𝑘 2 = 𝑛(𝑛+1)(2𝑛+1) 6 Now letting 𝑛 → ∞ in (1), we get that lim 𝑓(𝑛) = 𝑛→∞ 1 2 The product you are interested in is 𝑒 𝑓(𝑛) and since 𝑒 𝑥 is continuous, we have lim 𝑓(𝑛) 𝑒 𝑓(𝑛) = 𝑒 𝑛→∞ lim 𝑛→∞ = 𝑒 1/2 = √𝑒 Question #7 How to prove the following equality? ∑∞ 𝑛=1 1 ∏𝑚 𝑘=1(𝑛+𝑘) 1 = (𝑚−1)𝑚! Answer: ∞ ∞ ∞ 𝑛=1 1 𝑛=1 1 𝑛! 1 1 ∑ 𝑚 =∑ = ∑ ∏𝑘=1(𝑛 + 𝑘) (𝑛 + 𝑚)! 𝑚! ((𝑚 + 𝑛), (𝑛)) 𝑛=1 Now 1 ((𝑚+𝑛),(𝑛)) = 𝑚 ∫0 𝑑𝑥 𝑥 𝑚−1 (1 − 𝑥)𝑛 so that the sum in question is equal to, upon reversing sum and integral 1 1 1−𝑥 ∞ 𝑛 𝑚 ∫ 𝑑𝑥 𝑥 𝑚−1 ∑𝑛=1(1−𝑥) = 𝑚 ∫ 𝑑𝑥 𝑥 𝑚−1 𝑥 0 0 1 = 𝑚 ∫0 𝑑𝑥(𝑥 𝑚−2 − 𝑥 𝑚−1 ) 1 1 = 𝑚( − ) 𝑚−1 𝑚 1 = 𝑚−1 Putting this all together, ∑∞ 𝑛=1 ∏𝑚 1 𝑘=1(𝑛+𝑘) 1 = (𝑚−1)𝑚! as was to be shown. Question #8 Does the following series converge or diverge? ∑∞ 𝑛=1 1 √𝑛+√𝑛+1 Answer: It was not entirely obvious to me that the infinite sum of differences between square roots diverges, so I did telescoping: ∑∞ 𝑛=1 ( = = 1 √𝑛+1+√𝑛 ) = ∑∞ 𝑛=1 ( √𝑛+1−√𝑛 𝑛+1−𝑛 ) ∑∞ 𝑛=1(√𝑛 + 1 − √𝑛) lim ) 𝑁 𝑁→∞ ∑𝑛=1(√𝑛+1−√𝑛) = lim (√2 − √1 + √3 − √2 + ⋯ + √𝑁 + 1 − √𝑁) 𝑁→∞ = lim (√𝑁 + 1 − √1) 𝑁→∞ So the limit of partial sum is equal to infinity. Question #9 Prove that: lim (1 + 𝑛→∞ 1 𝑎𝑛 ) = 𝑒 𝑖𝑓 lim 𝑎𝑛 = ∞. 𝑛→∞ 𝑎𝑛 Answer: Given that lim 𝑏𝑛 = lim lim (1 + 𝑛→∞ 𝑛→∞ 1 𝑎𝑛 𝑎𝑛 ) 1 = 0 we have, 𝑛→∞ 𝑎𝑛 = lim 𝑒𝑥𝑝 ( 1 𝑛 ln(1+𝑎 ) 1 𝑎𝑛 𝑛→∞ ) = lim 𝑒𝑥𝑝 ( ln(1+ℎ) )=𝑒 ℎ ℎ→0 similarly 1 lim (1 + 𝑏𝑛 )𝑏𝑛 = lim 𝑒𝑥𝑝 ( 𝑛→∞ ln(1+𝑏𝑛) 𝑏𝑛 𝑛→∞ ) = lim 𝑒𝑥𝑝 ( ℎ→0 ln(1+ℎ) )=𝑒 ℎ Question #10 Evaluate ∑∞ 𝑛=1 1 using Parseval's theorem (Fourier series). 𝑛4 Answer: Let 𝑓(𝑥) = 𝑥 2 for 𝑥 ∈ (−𝜋, 𝜋). Computing the Fourier coefficients gives 𝑎𝑛 = 1 𝜋 ∫ 𝑥2 2𝜋 −𝜋 𝑒 𝑖𝑛𝑥 𝑑𝑥 = 2 cos(𝜋𝑛) 𝑛2 for 𝑛 ∈ 𝑍, 𝑛 ≠ 0 , and 𝑎0 = Therefore |𝑎𝑛 |2 = 4 𝑛4 =2 (−1)𝑛 𝜋 ∫ 𝑥2 2𝜋 −𝜋 1 𝑛2 𝑖𝑛𝑥 𝑒 𝑑𝑥 = 2 cos(𝜋𝑛) for 𝑛 ∈ 𝑍, 𝑛 ≠ 0 and |𝑎0 |2 = By Plancherel/Parseval's theorem, 𝜋4 9 + 8 ∑∞ 𝑛=1 1 𝑛4 2 = ∑∞ 𝑛=−∞|𝑎𝑛 | = Simplifying, this gives ∑∞ 𝑛=1 1 𝑛4 = 𝜋4 1 8 1 𝜋4 9 90 ( − )= 5 𝑛2 1 𝜋 ∫ 𝑥 4 𝑑𝑥 = 2𝜋 −𝜋 𝜋4 5 𝜋4 9 =2 (−1)𝑛 𝑛2