Questions and Answers #23 Series

Questions and Answers Sheet 23 Series Question #1 Determine the convergence of 1 𝑛 ∑+∞ 𝑛=1 ∫0 sin 𝑥 1+𝑥 𝑑𝑥 Answer: ∀𝑥 ≥ 0: sin 𝑥 ≤𝑥 1+𝑥 Because, for all 𝑥 ≥ 0 , 1 sin(𝑥) ≤ 𝑥 and ≤1 1+𝑥 So 1 ∫0𝑛 sin(𝑥) 1+𝑥 1 𝑑𝑥 ≤ ∫0𝑛 𝑥𝑑𝑥 = And ∑∞ 𝑛=1 1 1 2𝑛2 is convergent. 𝑛2 Question #2 How should I prove convergence using Cauchy sequences ∑𝑛𝑘=1 sin(𝑘 3 +1) (4𝑘+1)(4𝑘+5) Answer: You have sin(𝑘 2 +1) 𝑛+𝑝 |𝑥𝑛+𝑝 − 𝑥𝑛 | = |∑𝑘=𝑛+1 (4𝑘+1)(4𝑘+5)| sin(𝑘 2 +1) 𝑛+𝑝 ≤ ∑𝑘=𝑛+1 |(4𝑘+1)(4𝑘+5)| 1 1 1 𝑛+𝑝 = ∑𝑘=𝑛+1 ((4𝑘+1) − (4𝑘+5)) 4 1 = ( ≤ 1 4 4𝑛+5 1 − 1 4(𝑛+𝑝)+5 ) 4(4𝑛+5) and so choosing 𝑁 > 1 16𝜖 will provide an N for you to use in proving that [𝑥𝑛 ]1∞ is Cauchy. Question #3 Proof that Dirichlet series ∑∞ 𝑛=1 2𝑤(𝑛) 𝑛2 = 5 2 Answer: 2𝑤 = 1 ⋅ 𝜇 2 therefore ∑+∞ 𝑛=1 2𝑤(𝑛) = 𝜁(2) = = ∑+∞ 𝑛=1 𝑛2 𝜁(2) 𝜁(4) 𝜋2 𝜋2 /6 6 𝜋4 /90 1 𝑛2 ∑+∞ 𝑛=1 𝜇(𝑛)2 𝑛2 = 5 2 Indeed, +∞ 𝜁(2𝑠) ∑ 𝑛=1 1 = ∑+∞ 𝑝=1 +∞ 𝜇(𝑛)2 𝜇(𝑛)2 = ∑ 𝑠 (𝑛𝑚2 ) 𝑠 𝑛 𝑛,𝑚=1 = 𝜁(𝑠) 𝑝𝑠 because every integer 𝑝 ≥ 1 can be uniquely written as 𝑝 = 𝑛𝑚2 where 𝑚 ≥ 1 and 𝑛 ≥ 1 is a squarefree integer. Question #4 𝑒 −𝑥𝑛 ln 𝑥 Convergence of ∑∞ 𝑛=1 𝑥 2 +𝑛 if x>0 Answer: I appear to get a different answer. 𝑥 −𝑛𝑥 𝑎𝑛 = 𝑥 2 +𝑛 𝑥 −𝑥 1 ⇒ lim (𝑎𝑛 )𝑛 = lim 𝑛→∞ 𝑛→∞ 1 (𝑥 2 + 𝑛)𝑛 𝑥 −𝑥 = lim 1 𝑛→∞ 2 𝑛 𝑥 𝑛 (1+ 2 )𝑛 𝑥 1 2 Now 𝑥 𝑛 → 1 and (1 + 1 1 𝑛 𝑛 ) 𝑥2 𝑛 1 → 𝑒 𝑛𝑥2 = 𝑒 𝑥2 ⇒ lim (𝑎𝑛 )𝑛 = 𝑥 −𝑥 𝑒 𝑥2 𝑛→∞ Plugging this in a graphing calculator (or equivalently analyzing using derivatives) shows that its always lesser than 1. So, shouldn't the series converge. Question #5 Find convergence of: ∑∞ 𝑛=1 1 (𝑛!)2/𝑛 Answer: We have 𝑒𝑛 = 1 + 𝑛 1! + 𝑛2 2! + ⋯+ 𝑛𝑛 𝑛! +⋯> 𝑛𝑛 𝑛! Then 1 𝑛! < 𝑒𝑛 𝑛𝑛 so 1 𝑛!2/𝑛 < 𝑒2 𝑛2 and probably you can take it from here. Question #6 (−1)𝑛 (2𝑛)! What function does ∑∞ 𝑛=0 (1−2𝑛)𝑛!2 Answer: A good series to know offhand is 1 𝑛 = ∑∞ 𝑛=0((2𝑛), (𝑛))𝑡 √1−4𝑡 𝑥𝑛 4 𝑛 𝑦 2𝑛 correspond to? which almost gets us the series we desire with 𝑡 = − 1 = ∑∞ 𝑛=0((2𝑛), (𝑛)) (− 𝑥 √1+4𝑦2 𝑥 4𝑦 2 ) 𝑥 4𝑦 2 𝑛 except for the offending term 1-2n in the denominator. However a small ajustment to the original series with 𝑡 = 𝑧 −2 1 = 4 √1+ 2 𝑧 |𝑧| √𝑧 2 +4 𝑛 −2𝑛 = ∑∞ 𝑛=0((2𝑛), (𝑛))(−1) 𝑧 gives us the series we want via integration 𝑧 1−2𝑛 √𝑧 2 +4 √𝑧 2 + 4sgn(𝑧) = ∑∞ 𝑛=0((2𝑛), (𝑛)) 1−2𝑛 = |𝑧| 𝑧 −2𝑛 ∑∞ 𝑛=0((2𝑛), (𝑛)) 1−2𝑛 2𝑦 by substituting 𝑧 = √𝑥 1 𝑥 𝑛 ∑∞ 𝑛=0((2𝑛), (𝑛)) 1−2𝑛 (− 4𝑦 2 ) = √𝑦 2 +𝑥 |𝑦| = √1 + 𝑥 𝑦2 Question #7 Calculate: 4 𝑛+1 ∑∞ 𝑛=0 𝑛 ( ) 5 Answer: 4 𝑛+1 So, ∑𝑛≥0 𝑛 ( ) 5 4 2 4 3 4 4 5 5 5 = 1( ) + 2( ) + 3( ) + ⋯ Now, 4 2 4 3 4 4 𝑆 = 1 ( ) + 2 ( ) + 3 ( ) + ⋯ (1) 5 5 5 5𝑆 4 4 1 4 2 4 3 5 5 5 = 1( ) + 2( ) + 3( ) + ⋯ Subtract equation (1) from (2). 5𝑆 4 𝑆 4 4 4 2 4 3 5 5 4 5 4 2 4 3 5 5 5 − 𝑆 = + (2 − 1) ( ) + (3 − 2) ( ) + ⋯ = +( ) +( ) +⋯ This is nothing but infinite GP, where |r|<1 as 𝑟 = 4 5 So you get 𝑆 4 = 4 5 4 5 1− ⇒ 𝑆 = 16 Question #8 Proving that: ∑99 𝑛=1 √𝑛+1−√𝑛 2𝑛+1 < 9 20 Answer: As we known (2𝑛 + 1)2 = 4𝑛2 + 4𝑛 + 1 > 4𝑛2 + 4𝑛 = 4𝑛(𝑛 + 1), then 2𝑛 + 1 > 2√𝑛(𝑛 + 1), with 𝑛 ∈ 𝑁. Hence we have √𝑛+1−√𝑛 < ∑99 𝑛=1 2√𝑛(𝑛+1) 2𝑛+1 1 99 1 1 1 1 ∑ − = (1 − ) 2 𝑛=1 √𝑛 2 10 √𝑛+1 ∑99 𝑛=1 = √𝑛+1−√𝑛 = 9 20 Question #9 I have to prove the following: 1 𝑥 1/3 ∫0 1−𝑥 1 log 𝑥 9 𝑑𝑥 = ∑∞ 𝑛=0 (3𝑛+4)2 Answer: 1 𝑥 1/3 𝐼 = ∫0 1−𝑥 1 𝑥 1/3 1 log 𝑑𝑥 = − ∫0 𝑥 𝑑 | +𝑥 𝑑𝑡 𝑡=0 Note that 𝑡 1−𝑥 log(𝑥) 𝑑𝑥 = log(𝑥). Hence, by Fubuni-Tonelli we can interchange derivative/limit and integral sign: 1 𝑥 1/3 𝐼 = − ∫0 = 1−𝑥 𝑑 | + 𝑑𝑡 𝑡=0 1 𝑥 1/3 log(𝑥) 𝑑𝑥 = − ∫0 1 𝑥 1/3+𝑡 − ∫0 1 Now expand 𝐼= 1−𝑥 𝑑 ( |𝑡=0+ 𝑥 𝑡 ) 𝑑𝑥 1−𝑥 𝑑𝑡 𝑑𝑥 : 1−𝑥 ∞ 𝑛 1 |𝑡=0+ − ∫0 𝑥 1/3+𝑡 ∑𝑛=0 𝑥 𝑑𝑡 𝑑 𝑑𝑥 Again, by Fubuni-Tonelli: 1 1/3+𝑡+𝑛 𝑑 𝐼 = |𝑡=0+ − ∑∞ 𝑑𝑥 = 𝑛=0 ∫0 𝑥 𝑑𝑡 𝑑 −1 𝑑𝑡 𝑛+𝑡+ = ∑∞ 𝑛=0 [ |𝑡=0+ 4 3 1 = ∑∞ 𝑛=0 [ lim+ 2 𝑡→0 (𝑛+𝑡+4) 𝑑 | + 𝑑𝑡 𝑡=0 − ∑∞ 𝑛=0 1 4 3 𝑛+𝑡+ ] ] 3 = ∑∞ 𝑛=0 1 4 2 3 (𝑛+ ) 9 = ∑∞ 𝑛=0 (3𝑛+4)2 Question #10 How to prove: 𝑛!𝑛! 𝜋 2 = 18 ∑∞ 𝑛=0 (2𝑛+2)! Answer: One has 𝑆 = ∑∞ 𝑛=0 𝑛!𝑛! 1 (2𝑛+2)! = ∑∞ 𝑛=0 (2𝑛+2)(2𝑛+1)((2𝑛),(𝑛)) How you have the following Maclaurin expansion: arcsin 𝑥 √1−𝑥 2 22𝑛𝑥 2𝑛+1 = ∑∞ 𝑛=0 (2𝑛+1)((2𝑛),(𝑛)) so 𝑥 1 arcsin(2) 𝑆 = 2 ∫0 2 √1−(𝑥) 2 𝑥 1 𝜋 2 𝜋2 0 6 18 𝑑𝑥 = 2 [arcsin2 ( )] = 2 × ( ) = and you are done. 2