Questions and Answers #15 Series

Questions and Answers Sheet 15 Series Question #1 Define the sequence [𝑎𝑛 ]𝑛≥2 by 𝑎𝑛 = √1 + √2 + ⋯ + √𝑛 It is well known 𝑎𝑛 → 𝑙 for a certain real number l. Show that lim 𝑛→∞√𝑛 √𝑒 2 ⋅ 𝑛√𝑙 − 𝑎𝑛 = Answer: 1) We want to estimate the difference 𝑙 − 𝑥𝑛 . To do this note that 𝑙 − 𝑥𝑛 = ∑𝑚≥𝑛(𝑥𝑚+1 − 𝑥𝑚 ) So we find asymptotics for 𝑥𝑚+1 − 𝑥𝑚 2) Show that xn +1 − xn = n +1  (xn( k+1) + xn( k ) 1 k  n (𝑘) where 𝑥𝑛 = √𝑘 + √(𝑘 + 1) + ⋯ + √𝑛 (𝑘) (𝑘+1) 3) Using 𝑥𝑛 = √𝑘 + 𝑥𝑛 show that (𝑘) √𝑘\𝑙𝑒𝑞𝑥𝑛 ≤ √𝑘 + 1 (3.1) (𝑘) 1 1 2 √𝑘 𝑥𝑛 = √𝑘 + + 𝑂 ( ) (3.2) (𝑘) 𝑥𝑛 = √𝑘𝑒𝑥𝑝 ( 1 2√𝑘 +𝑂( 1 𝑘 3/2 )) (3.3) for 1 ≤ 𝑘 < 𝑛 What is more, constant in O symbols are independent of k and n 4) Using 3.3 show that 𝑥𝑛+1 − 𝑥𝑛 = hence 𝑥𝑛+1 − 𝑥𝑛 ∼ 1+𝑜(1) 2𝑛√(𝑛−1)! 𝑐𝑜𝑛𝑠𝑡 2𝑛√(𝑛−1)! 𝑒𝑥𝑝 (− ∑1≤𝑘<𝑛 ( 1 2√𝑘 +𝑂( 1 𝑘 3/2 ))) 𝑒𝑥𝑝(−√𝑛) and 𝑙 − 𝑥𝑛 ∼ 𝑥𝑛+1 − 𝑥𝑛 ∼ 𝑐𝑜𝑛𝑠𝑡 2𝑛√(𝑛−1)! 𝑒𝑥𝑝(−√𝑛) 5) Applying Stirling's formula we get 1 𝑛 𝑒 √𝑙 − 𝑥𝑛 ∼ 2 √𝑛 and the result follows. Question #2 1 How to calculate ∫0 𝑥 𝑥 𝑑𝑥 using series? I read from a book that 1 1 1 1 ∫0 𝑥 𝑥 𝑑𝑥 = 1 − 22 + 33 + ⋯ + (−1)𝑛 (𝑛+1)𝑛+1 + ⋯ Answer: Just write 𝑥 𝑥 = 𝑒 𝑥 ln 𝑥 = ∑∞ 𝑛=0 (𝑥 ln 𝑥)𝑛 𝑛! and use that 1 ∞ 𝑛 −(𝑛+1)𝑦 𝑦 𝑒 ∫ (𝑥 ln 𝑥)𝑛 𝑑𝑥 = (−1)𝑛 ∫0 𝑑𝑦, 0 To show the last formula, make the change of variables 𝑥 = 𝑒 −𝑦 so that 1 ∞ 𝑛 −(𝑛+1)𝑦 𝑦 𝑒 ∫ (𝑥 ln 𝑥)𝑛 𝑑𝑥 = (−1)𝑛 ∫0 𝑑𝑢, 0 which is clearly expressible in terms of the Gamma function. Question #3 I want to speed up the convergence of a series involving rational expressions the expression is 𝑥 ∑∞ 𝑥=1(−1) −𝑥 2 −2𝑥+1 𝑥 4 +2𝑥 2 +1 Answer: Here is the first step toward making the series evaluate quicker. You may need just as many terms but it will be faster. Break the series up through partial fractions. −𝑥 2 −2𝑥+1 𝑥 4 +2𝑥 2 +1 2(𝑥−1) = − (𝑥 2 +1)2 − 1 𝑥 2 +1 I will focus on the second term in the sum. ∑∞ 𝑥=0 (−1)𝑥 𝑥 2 +1 = ∑∞ 𝑛=0 1 4𝑛2 +1 1 − ∑∞ 𝑘=0 (2𝑘+1)2 +1 Each of the two sums can be evaluated by contour methods. In short, for a sufficient rational function f(z), lim ) 𝑘=𝑁 𝑁→∞ ∑𝑘=−𝑁 𝑓(𝑘) is equal to the negative of the sum of the residues of 𝜋𝑓(𝑧) cot(𝜋𝑧) at the poles of f(z) . We can use this because we are summing an even function. This yields ∑∞ 𝑛=0 1 1 𝜋 = (2 + 𝜋 coth ( )) 4𝑛2 +1 1 4 ∑∞ 𝑘=0 (2𝑘+1)2 +1 2 1 𝜋 4 2 = 𝜋 tanh ( ) Subtract these (and simplify) to find ∑∞ 𝑛=0 (−1)𝑥 𝑥 2 +1 1 = (𝜋 𝑐𝑠𝑐 ℎ(𝜋) − 1) 2 Now we can conclude our first step by writing... 𝑥 ∑∞ 𝑥=1(−1) −𝑥 2 −2𝑥+1 𝑥 4 +2𝑥 2 +1 1 2(𝑥−1) 2 +1)2 𝑥 = − (𝜋 𝑐𝑠𝑐 ℎ(𝜋) − 1) − ∑∞ 𝑥=1(−1) (𝑥 2 It may be tempting to try the same approach for the remaining series but we are not summing an even function so it will not be as easy. Question #4 Find −1 𝑀 ≔ ∑∞ 𝑛=1 tan 2 𝑛2 Answer: −1 ∑∞ 𝑛=1 tan 2 𝑛2 ∞ −1 (1+𝑛)+(1−𝑛) ∑ = 𝑛=1 tan 1−(1+𝑛)(1−𝑛) −1 −1 = ∑∞ 𝑛=1 tan (1 + 𝑛) + tan (1 − 𝑛) ∞ −1 = ∑𝑛=1 tan (𝑛 + 1) − tan−1 (𝑛 − 1) Edit: ∑∞ 𝑛=1 was missing in your question before edit. I am not going to delete this. However your proof is now correct. −1 𝑀 = lim (tan−1 (𝑚 + 1) + tan−1 𝑚 − tan−1 1 − tantan 𝜋 𝑚→∞ 𝜋 𝜋 𝜋 3𝜋 4 4 = + − −0=𝜋− = 2 2 4 0) Question #5 2 𝑛 If 𝑎𝑛 = (1 + ) , then find 𝑛 lim (1 − 𝑎𝑛 𝑛 𝑛 𝑛→∞ ) Answer: Due to (1 − 𝑛 𝑎𝑛 𝑛 𝑛 ) = [(1 − lim (1 − 𝑛 𝑎𝑛 −𝑎𝑛 𝑛→∞ 𝑛 ) 𝑎𝑛 −𝑎𝑛 ) ] 𝑛 −𝑎ℎ 𝑛 𝑛 𝑛 = [(1 − 𝑎𝑛 −𝑎𝑛 ) ] 𝑛 −𝑎𝑛 =𝑒 and lim (−𝑎𝑛 ) = 𝑒 2 𝑛→∞ 𝑛 Let 𝐴𝑛 = (1 − 𝑎𝑛 −𝑎𝑛 ) 𝑛 and 𝐵𝑛 = −𝑎𝑛 , by the ''claim'' below, you can get the result! Question #6 Prove that: ∑∞ 𝑛=1 𝑛2 (𝑛−1) 2𝑛 = 20 Answer: Note that ∑𝑛 𝑡 𝑛 = ∑ 𝑛(𝑛 − 1)𝑡 𝑛−2 𝑛 1 for every |t|<1 hence, differentiating this twice and three times, 2 = , (1 − 𝑡)3 1−𝑡 6 ∑𝑛 𝑛(𝑛 − 1)(𝑛 − 2)𝑡 𝑛−3 = (1−𝑡)4 1 For 𝑡 = , this reads 2 ∑𝑛 𝑛(𝑛 − 1) 1 2𝑛−2 = 2 ⋅ 23 ∑𝑛 𝑛(𝑛 − 1)(𝑛 − 2) 1 2𝑛−3 = 6 ⋅ 24 which implies 1 1 ∑𝑛 𝑛(𝑛 − 1) 𝑛 = ⋅ 2 ⋅ 23 = 4 2 4 ∑𝑛 𝑛(𝑛 − 1)(𝑛 − 2) 1 2𝑛 1 = ⋅ 6 ⋅ 24 = 12 8 Finally, 𝑛2 (𝑛 − 1) = 2 ⋅ 𝑛(𝑛 − 1) + 𝑛(𝑛 − 1)(𝑛 − 2) hence 1 1 ∑𝑛 𝑛2 (𝑛 − 1) 𝑛 = 2 ⋅ ∑𝑛 𝑛(𝑛 − 1) 𝑛 + ∑𝑛 𝑛(𝑛 − 1)(𝑛 − 2) 2 2 1 2𝑛 = 2 ⋅ 4 + 12 This approach can be made shorter if one notices once and for all that, for every nonnegative k, 1 ∑𝑛 𝑛(𝑛 − 1) … (𝑛 − 𝑘) 𝑛 = 2 ⋅ (𝑘 + 1)! 2 Question #7 Evaluate ∑𝑛,𝑘 1 ((𝑛),(𝑘)) where the summation ranges over all positive integers n, k with 1 < 𝑘 < 𝑛 − 1 Answer: Let l=n-k , we have −1 𝑛−2 ∑∞ 𝑛=4 ∑𝑘=2 ((𝑛), (𝑘)) ∞ = ∑∞ 𝑘=2 ∑𝑛=𝑘+2((𝑛), (𝑘)) ∞ = ∑∞ 𝑘=2 ∑𝑙=2((𝑘 + 𝑙), (𝑘)) ∞ = ∑∞ 𝑘=2 ∑𝑙=2(𝑘 + 𝑙 + 1) = ∞ ∑∞ 𝑘=2 ∑𝑙=2(𝑘 + 𝑙 1 ∞ ∫0 ∑∞ 𝑘=2 ∑𝑙=2[(𝑘 + −1 −1 𝑘!𝑙! ∞ = ∑∞ 𝑘=2 ∑𝑙=2 (𝑘+𝑙)! Γ(𝑘+1)Γ(𝑙+1) Γ(𝑘+𝑙+2) 1 𝑘 1) ∫0 𝑡 (1 − 𝑘 (1 𝑡)𝑙 𝑑𝑡 = + 𝑙 + 1)𝑡 − 𝑡)𝑙 ]𝑑𝑡 Notice when s and t are independent, we have ∞ 𝑘 ∑∞ 𝑘=2 ∑𝑙=2(𝑘 + 𝑙 + 1)𝑡 ∞ 𝑠 𝑙 = ∑∞ 𝑘=2 ∑𝑙=2 (𝑡 = (𝑡 = 𝑑 +𝑠 𝑑 𝑑 𝑑𝑡 +𝑠 𝑑 𝑑𝑠 + 1) 𝑡 𝑘 𝑠 𝑙 𝑡 2𝑠2 + 1) (1−𝑡)(1−𝑠) 𝑑𝑡 𝑑𝑠 𝑠 2 𝑡 2 (5−4(𝑠+𝑡)+3𝑠𝑡) (1−𝑠)2 (1−𝑡)2 Substitute s by 1-t, we obtain ∞ 𝑘 𝑙 ∑∞ 𝑘=2 ∑𝑙=2(𝑘 + 𝑙 + 1)𝑡 (1 − 𝑡) = 1 + 3𝑡(1 − 𝑡) As a result, ∞ ∑∞ 𝑘=2 ∑𝑙=2((𝑛), (𝑘)) −1 1 = ∫0 (1 + 3𝑡(1 − 𝑡))𝑑𝑡 = Question #8 What is the value of ∑∞ 𝑘=1 1 (3𝑘+1)(3𝑘+2) Answer: I like solution with Gamma/Beta function: 1 Γ(3𝑘+1) = ∑+∞ 𝑘=1 Γ(3𝑘+3) (3𝑘+1)(3𝑘+2) ∑+∞ 𝑘=1 𝐵(3𝑘 + 1,2) 1 3𝑘 ∑+∞ 𝑘=1 ∫0 𝑥 (1 − 𝑥)𝑑𝑥 1 3𝑘 ∫0 ∑+∞ 𝑘=1 𝑥 (1 − 𝑥)𝑑𝑥 3 1 𝑥 ∫0 1+𝑥+𝑥 2 𝑑𝑥 1 √3𝜋 ∑+∞ 𝑘=1 = = = = = 9 − 2 Question #9 How to prove: ∑∞ 𝑘=1 𝑘−1 2𝑘(1+𝑘)(1+2𝑘) = log 𝑒 8 − 2 3 2 Answer: Let me try. Note that the 𝑅𝐻𝑆 = 3 ln 2 + 2 = 2 + 3 ∑∞ 𝑘=1 (−1)𝑘+1 = 2 + 3 ∑∞ 𝑘=1 𝑘 1 2𝑘−1 − 1 2𝑘 We have 𝑘−1 ∑∞ 𝑘=1 = 2 = ∑∞ 𝑘=1 ( =2+ 2𝑘−(𝑘+1) = ∑∞ 𝑘=1 2𝑘(𝑘+1)(2𝑘+1) 1 ∑∞ 𝑘=1 ((𝑘+1)(2𝑘+1) 1 − − − 2𝑘(𝑘+1)(2𝑘+1) 1 ) 2𝑘(2𝑘+1) 1 1 2𝑘+1 𝑘+1 2𝑘 1 1 ∞ 3 ∑𝑘=1 − 2𝑘−1 2𝑘 + 2𝑘+1 ) =RHS Question #10 I try to solve the following sum: ∞ ∑∞ 𝑘=1 ∑𝑛=1 1 𝑘 2 𝑛+2𝑛𝑘+𝑛2 𝑘 Answer: Let's write: +∞ 𝑓(𝑥) = ∑+∞ 𝑘=1 ∑𝑛=1 𝑥 𝑛+𝑘+2 𝑛𝑘(𝑛+𝑘+2) then: +∞ 𝑓 ′ (𝑥) = ∑+∞ 𝑘=1 ∑𝑛=1 𝑥 𝑛+𝑘+1 𝑛𝑘 = − ∑+∞ 𝑘=1 𝑥 𝑘+1 ln(1−𝑥) 𝑘 We want to know the value of f(1), so we integrate: +∞ +∞ 1 1 𝐻𝑘+2 𝑓(1) = − ∑ ∫ 𝑥 𝑘+1 ln(1 − 𝑥) 𝑑𝑥 = ∑ 𝑘 0 𝑘(𝑘 + 2) 𝑘=1 1 𝐻𝑘+2 2 𝑘 = ∑+∞ 𝑘=1 ( 𝑘=1 − 𝐻𝑘+2 𝑘+2 1 2 𝑘 ) = ∑+∞ 𝑘=1 ( 1 1 3 2 2 2𝑘 = (𝐻1 + 𝐻2 + ∑+∞ 𝑘=1 ( 1 1 𝐻𝑘 +𝑘+1+𝑘+2 − 1 1+𝑘 − 1 − )) = 2(𝑘+2) 𝐻𝑘+2 𝑘+2 ) 7 4 The very last sum telescopes: ∑+∞ 𝑘=1 ( 3 2𝑘 − 1 1+𝑘 − 1 1 1 𝑘 1+𝑘 ) = ∑+∞ 𝑘=1 (( − 2(𝑘+2) )+( 1 2𝑘 − 1 )) = 2(𝑘+2) 7 4 And the integral I've used is obtained by integrating by parts: 1 1 𝑘+2 𝑥 𝑘+2 ln(1 − 𝑥) 1 1 𝑥 −1+1 |0 + ∫ 𝑥 𝑘+1 ln(1 − 𝑥) 𝑑𝑥 = ∫ 𝑑𝑥 𝑘+2 𝑘+2 0 1−𝑥 0 −= 𝐻𝑘+2 𝑘+2 𝜖 𝑑𝑥 + lim− (𝑥 𝑘+2 ln(1 − 𝑥) + ∫0 𝜖→1 1−𝑥 )=− 𝐻𝑘+2 𝑘+2