Questions and Answers #21 Series

Questions and Answers Sheet 21 Series Question #1 When does this series: ∑∞ 𝑛=0 sin(𝑛! 𝜋𝑥) converge? Answer: I think I can prove it for the first one, and then the others should be similar. In the case of ∑∞ 𝑛=0 sin(𝑛! 𝜋𝑒) we need to consider the value of 𝑛! 𝜋𝑒 mod 𝜋, and it's enough to consider the value of 𝑛! 𝑒 mod. Now 1 𝑛! 𝑒 = 𝑛! ∑∞ 𝑘=0 𝑘! so that 𝑛! 𝑒 mod 1 = 𝑛! ∑∞ 𝑘=𝑛+1 Question #2 That is equivalent to 𝜋 = √12 ∑ 𝑛 ≥ 0 (−1)𝑛 3𝑛(2𝑛+1) Answer: Another answer: 𝜋 1 √12 = √3 arctan ( ) √3 1 = √3 ∫0√3 1 √3 1 1+𝑡 2 𝑑𝑡 2 𝑛 = √3 ∫0 ∑∞ 𝑛=0(−𝑡 ) 𝑑𝑡 1 √3 𝑡 2𝑛 𝑛∫ = √3 ∑∞ 𝑛=0(−1) 0 𝑛 = √3 ∑∞ 𝑛=0(−1) [ = √3 ∑∞ 𝑛=0 (−1)𝑛 √3 = ∑∞ 𝑛=0 3𝑛 (2𝑛+1) ∑∞ 𝑛=0 𝑑𝑡 1 √3 ] 2𝑛+1 0 1 2𝑛+1 (( ) 2𝑛+1 √3 (−1)𝑛 = √3 𝑡 2𝑛+1 2𝑛 (√3) (−1)𝑛 (2𝑛+1) So 𝜋 = √12 ∑∞ 𝑛=0 (−1)𝑛 3𝑛(2𝑛+1) Question #3 Prove that: ∑∞ 𝑛=1 (−1)𝑛+1 𝑛(𝑛+1) = 2 ln 2 − 1 − 0) 1 𝑘! 1 < ∑∞ 𝑘=1 (𝑛+1)𝑘 = 1 𝑛 → 0 𝑎𝑠 𝑛 → ∞. Answer: ∑𝑛𝑘=1 = (−1)𝑘+1 1 1 𝑘 𝑘+1 = ∑𝑛𝑘=1(−1)𝑘+1 ( − 𝑘(𝑘+1) (−1)𝑘+1 2 ∑𝑛𝑘=1 𝑘 ) = ∑𝑛𝑘=1 (−1)𝑘+1 𝑘 + ∑𝑛𝑘=1 (−1)𝑘+2 𝑘+1 (−1)𝑛+1 −1− 𝑛+1 Hence, ∑∞ 𝑛=1 (−1)𝑛+1 = 2 ∑∞ 𝑛=1 𝑛(𝑛+1) (−1)𝑛+1 𝑛 − 1 = 2 ln 2 − 1 since ∑𝑛𝑛=1 (−1)𝑛+1 = ln 2 𝑛 Proof. If |x|<1 , then 1 1+𝑥 = ∑𝑛𝑘=0(−1)𝑘 𝑥 𝑘 + (−1)𝑛+1 𝑥 𝑛+1 1+𝑥 and hence 𝑛 𝑥 (−1)𝑛+1 𝑛+1 𝑡 𝑘 𝑑𝑡 𝑡 𝑑𝑡 log(1 + 𝑥) = ∫ = ∑(−1)𝑘 ∫0 𝑡 𝑑𝑡 + ∫ 1 + 𝑡 1 + 𝑡 0 0 𝑥 𝑘=0 = ∑𝑛𝑘=0 (−1)𝑘 𝑥 𝑛+1 𝑛+1 + 𝑅𝑛 (𝑥) Clearly, for 𝑥 ∈ [0,1] 𝑥 𝑡 𝑛+1 𝑑𝑡 |𝑅𝑛 (𝑥)| = ∫0 1+𝑡 𝑥 ≤ ∫0 𝑡 𝑛+1 𝑑𝑡 ≤ 1 𝑛+2 Hence ln 2 = lim− ln(1 + 𝑥) = 𝑥→1 lim (−1)𝑘 𝑥𝑛+1 𝑥→1− ∑𝑛 𝑘=0 𝑛+1 + lim− 𝑅𝑛 (𝑥) = ∑𝑛𝑘=0 𝑥→1 (−1)𝑘 𝑛+1 + 𝑅𝑛 (1) and hence ∑∞ 𝑛=1 (−1)𝑛+1 𝑛 = lim 𝑛→∞ ∑𝑛 𝑘=0 = lim (ln 2 + 𝑅𝑛 (1)) = ln 2 (−1)𝑘 𝑛+1 𝑛→∞ Question #4 I want to evaluate the sum ∑∞ 𝑛=2 𝑛4 +3𝑛2 +10𝑛+10 2𝑛(𝑛4 +4) Answer: So once you have obtained 1 −1 ( 2𝑛 𝑛2 +2𝑛+2 + 4 𝑛2 −2𝑛+2 + 1) then observe the following: 𝑛2 + 2𝑛 + 2 = (𝑛 + 1)2 + 1 and 𝑛2 − 2𝑛 + 2 = (𝑛 − 1)2 + 1. Which means the given sum becomes ∑∞ 𝑛=2 [ −1 2𝑛 [(𝑛+1)2 +1] + where 𝑢𝑛 ≔ If you 1 ] + ∑∞ 𝑛=2 2𝑛−2 [(𝑛−1)2 +1] 1 2𝑛−2 [(𝑛−1)2 +1] calculate ∑∞ 𝑛=2[𝑢𝑛 − 1 2𝑛 = ∑∞ 𝑛=2[𝑢𝑛 − 𝑢𝑛+2 ] + 1 1 2 10 𝑢𝑛+2 ] = 𝑢2 + 𝑢3 = + 1 1 2 10 Hence the final sum is 2 × × 1 2 =⋯ Question #5 Does this double series has closed form (i.e. can be computed) ? ∞ ∑∞ 𝑚=1 ∑𝑛=1 1 10𝑛𝑚 Answer: Trivially ∞ ∑∞ 𝑚=1 ∑𝑛=1 1 10𝑚𝑛 = ∑∞ 𝑚=1 1 10𝑚 −1 = − ∑∞ 𝑚=1 1 1−10𝑚 and the last sum can be computed, using the identity ∑∞ 𝑚=1 1 1−𝑎𝑚 𝜓1/𝑎 (1)+log(𝑎−1)+log(1/𝑎) = log(𝑎) where 𝜓𝑞 (𝑧) = − log(1 − 𝑞) + log(𝑞) ∑∞ 𝑚=0 𝑞 𝑛+𝑧 1−𝑞 𝑛+𝑧 is the q-Polygamma function. Hence ∞ ∑∞ 𝑚=1 ∑𝑛=1 1 10𝑚𝑛 =− 𝜓1/10 (1)+log(9)+log(1/10) log(10) ≈ 0.122324 Question #6 Prove the converges of ∑∞ 𝑛=1 √2𝑛−1 ln(4𝑛+1) 𝑛(𝑛+1) Answer: I would suggest you start with some rough hand-wavy approximations and then make it rigorous. For example, when n large, √2𝑛−1 𝑛(𝑛+1) √2𝑛 ≈ 𝑛2 = √2𝑛−3/2 and ln(4𝑛 + 1) ∼ ln 𝑛 Let's try integrating ∫ 𝑥 −3/2 ln 𝑥 𝑑𝑥 1 Let 𝑢 = ln 𝑥 , 𝑑𝑢 = 𝑑𝑥, 𝑑𝑣 = 𝑥 −3/2 , 𝑣 = −2𝑥 −1/2 . This gives an integral of 𝑥 −2𝑥 −1/2 ln 𝑥 − ∫ −2𝑥 −3/2 𝑑𝑥 = −2𝑥 −1/2 ln 𝑥 − 4𝑥 −1/2 + 𝐶 = −2 ln 𝑥 √𝑥 −4 1 √𝑥 +𝐶 So the integral from, say, 1 to +∞ of this thing is quite clearly finite. Question #7 I'm trying to evaluate: 𝑛 ∑∞ 𝑛=0(−1) ln 2+2𝑛 1+2𝑛 Answer: We have the series expansion of the digamma function: 1 1 𝜓(𝑧) = −𝛾 + ∑∞ − 𝑛=0 𝑛+1 𝑛+𝑧 Integrating both sides gives ln Γ (𝑧) = (1 − 𝑧)𝛾 + ∑∞ 𝑛=0 𝑧−1 𝑛+1 𝑛+1 + ln [ 𝑛+𝑧 ] Hence we have (𝑛+1/2)(𝑛+3/4) ln Γ (1) + ln Γ (1/4) − ln Γ (1/2) − ln Γ (3/4) = ∑∞ 𝑛=0 ln [ Γ(1/4) 1 Γ(3/4) 2 ln [ (4𝑛+2)(4𝑛+3) ] − ln(𝜋) = ∑∞ 𝑛=0 ln [(4𝑛+4)(4𝑛+1)] (𝑛+1)(𝑛+1/4) ] Question #8 2 2 𝑛 Does the series ∑∞ 𝑛=1 sin (2𝜋√𝑛 + 𝑎 sin 𝑛 + (−1) ) converge? Answer: This series is convergent. As n tends to +∞, we may write 𝑢𝑛 ≔ sin (2𝜋√𝑛2 + 𝑎2 sin 𝑛 + (−1)𝑛 ) = sin (2𝜋𝑛√1 + = sin (2𝜋𝑛 (1 + = sin (2𝜋𝑛 + = sin ( 𝑎2 sin 𝑛 + 2𝑛2 𝑛 + + 𝑛2 𝜋𝑎2 sin 𝑛 𝜋𝑎2 sin 𝑛 𝑛 𝑎2 sin 𝑛 𝑛 𝑛2 ) (−1)𝑛 2𝑛2 𝜋(−1)𝑛 + 𝜋(−1)𝑛 (−1)𝑛 𝑛 1 + 𝑂 ( 4))) 𝑛 1 + 𝑂 ( 3)) 𝑛 1 + 𝑂 ( 3)) 𝑛 𝜋𝑎2 sin 𝑛 𝜋(−1)𝑛 1 + + 𝑂 ( 3) 𝑛 𝑛 𝑛 sin 𝑛 Now recall that ∑+∞ is convergent, moreover 𝑛=1 = 𝑛 sin 𝑛 ∑+∞ 𝑛=1 𝑛 = 𝑖𝑛 𝑒 𝐼 ∑+∞ 𝑛=1 𝑛 = 𝐼(− log(1 − 𝑒 𝑖 )) = 𝜋−1 2 Then it is clear that your initial series ∑+∞ 𝑛=1 𝑢𝑛 is convergent, being the sum of convergent series. Question #9 Show that: ∞ 2𝑚−1 ∑ ∑ 𝑚=1 𝑛=2 𝑚 = 1 − 𝛾. (2𝑛 + 1)(2𝑛 + 2) 𝑚−1 Answer: ∑∞ 𝑚=1 ∑𝑛=2𝑚 −1 𝑚 ( 1 2𝑛+1 lim𝑚 𝑚+1 2 −1 𝑁→∞ ∑𝑁 𝑚=1(∑𝑛=2𝑚 = 𝑛+1 − 𝑛=2 lim𝑚 𝑁+1 = lim (∑2𝑛=2𝑁−1 𝑁→∞ 𝑁+1 𝑛+1 ) 𝑚 𝑚 ) −∑2 −1 𝑚−1 ) 𝑛+1 ) 𝑚+1 2 −1 𝑁→∞ ∑𝑛 𝑚=1(∑𝑛=2𝑚 1 2𝑛+2 𝑛+1 𝑚 𝑚 ) −∑2 −1 𝑚−1 1 𝑛=2 𝑛+1 𝑁+1 −1 1 − − ∑2𝑛=2 2 𝑛+1 ) 𝑁+1 1 = lim ((𝑁 + 1)(log(2) + 𝑂(2−𝑁 )) + 1 − ∑2𝑛=1 𝑁→∞ 𝑛 ) = 1−𝛾 Question #10 Consider the series 2 2 ∑_{𝑛 = 1}^∞(𝑒 −𝑛 − 𝑒 −(𝑛+𝑥) Answer: This is not a complete solution but a collection of interesting partial results. Let the sum in question be 2 2 𝑓(𝑥) = ∑∞ 𝑛=1(𝑒𝑥𝑝(−𝑛 ) − 𝑒𝑥𝑝(−(𝑛 + 𝑥) )) Symmetry: Define 1 1 2 2 𝑔(𝑥) = 𝑓 (𝑥 − ) + then g(x) is antisymmetric: g(-x)=-g(x) Hence follows f(-x)=-1-f(x-1) (1) 2) Asymptotic values 1 1 2 𝑒 2 𝑎 = 𝑓(𝑥 → +∞) = ∑∞ 𝑛=1 𝑒𝑥𝑝(−𝑛 ) = (𝑣3 (0, ) − 1) ≅ 0.386319 where 𝑣3 is a thetea function. And, using (4) 𝑓(𝑥 → −∞) = −1 − 𝑎 ≅ −1.386319 3) Integer values of x lead to finte sums due to cancellations f(0)=0 𝑓(1) = 𝑒𝑥𝑝(−12 ) + 𝑒𝑥𝑝(−22 ) + 𝑒𝑥𝑝(−32 ) + ⋯ − 𝑒𝑥𝑝(−(1 + 1)2 ) = 𝑒𝑥𝑝(−12 ) 𝑓(2) = 𝑒𝑥𝑝(−12 ) + 𝑒𝑥𝑝(−22 ) and generally, 𝑓(𝑘) = ∑𝑘𝑛=1 𝑒𝑥𝑝(−𝑛2 ) , 𝑘 = 1,2, … and by (1) 2 𝑓(−𝑘) = −1 − 𝑓(𝑘 − 1) = −1 − ∑𝑘−1 𝑛=1 𝑒𝑥𝑝(−𝑛 ) , 𝑘 = 1,2, …