Questions and Answers #13 Series

Questions and Answers Sheet 13 Series Question #1 Find (𝑛+1)√𝑛+1 lim 𝑛√𝑛 𝑛→∞ Answer: Notice that for 𝑓(𝑥) = √𝑥 ln 𝑥 you have 𝑓 ′ (𝑥) = ln 𝑥 2√𝑥 − 1 √𝑥 Now by mean value theorem 𝑓(𝑛 + 1) − 𝑓(𝑛) = 𝑓 ′ (𝜃𝑛 ) for some 𝜃𝑛 such that 𝑛 < 𝜃𝑛 < 𝑛 + 1 (by Mean Value Theorem). This means that for 𝑛 → ∞ we have 𝜃𝑛 → ∞ If we notice that lim 𝑓 ′ (𝑥) = 0 we get that 𝑥→∞ lim (√𝑛 + 1 ln(𝑛 + 1) − √𝑛 ln 𝑛) = lim 𝑓 ′ (𝜃𝑛 ) = 0 𝑛→∞ 𝑛→∞ Question #2 Prove the following inequality for every 𝑛 ≥ 1: ∑𝑛𝑘=1 1 𝑘 2 +3𝑘+1 ≤ 13 20 Answer: Since ∑𝑛𝑘=1 < = 1 5 1 5 1 1 is monotone decreasing for 𝑘 ≥ 0 , we have 𝑘 2 +3𝑘+1 1 1 ≤ + 1 +∫ ∞ 1 2 𝑘 2 +3𝑘+1 𝑘 2 +3𝑘+1 5 11 ∞ 1 1 + + ∫2 2 𝑑𝑘 11 𝑘 +2𝑘+1 1 −1 ∞ | + + 11 𝑘+1 2 1 1 13 = + 5 11 + < 3 𝑑𝑘 20 1. I remembered that monotonic series can be bounded by integrals, by thinking of the series as a right Riemann sum. This suggests I try the bound ∞ 1 1 1 ∑𝑛𝑘=1 2 ≤ + ∫1 2 𝑑𝑘 𝑘 +3𝑘+1 5 𝑘 +3𝑘+1 2. That integral on the right looks mighty unpleasant; the denominator doesn't factor so the antiderivative will have logs and arctans galore. But I can bound the integral by the much nicer perfect square 𝑛 ∑ ∞ 1 1 1 1 1 ≤ + ∫ 𝑑𝑘 = + 2 2 𝑘 + 3𝑘 + 1 5 5 2 1 𝑘 + 2𝑘 + 1 𝑘=1 14 = 20 3. The bound is barely not tight enough. Pulling out more terms from the sum should improve it, so I try ∞ 1 1 1 1 ∑𝑛𝑘=1 2 ≤ + + ∫2 2 𝑑𝑘 𝑘 +3𝑘+1 5 11 𝑘 +2𝑘+1 which after working out the details turns out to work. Question #3 What tools would you gladly recommend me for computing precisely the limit below? Maybe a starting point? 1 lim 𝑛→∞ log(𝑛) ∑𝑛𝑘=1 cos(sin(2𝜋 log(𝑘))) 𝑘 Answer: Let 𝑓(𝑥) = 𝑓 ′ (𝑥) = − cos(sin(2𝜋𝑥)) , we have 𝑥 2𝜋 cos(2𝜋𝑥) sin(sin(2𝜋𝑥))+cos(sin(2𝜋𝑥)) 𝑥2 ⇒ |𝑓 ′ (𝑥)| ≤ 2𝜋+1 𝑥2 By MVT, for any 𝑥 ∈ (𝑘, 𝑘 + 1] we can find a 𝜁 ∈ (0,1) such that 2𝜋+1 |𝑓(𝑥) − 𝑓(𝑘)| = |𝑓 ′ (𝑘 + 𝜁(𝑥 − 𝑘))|(𝑥 − 𝑘) ≤ 2 𝑘 This implies 𝑘+1 |∫𝑘 𝑓(𝑥)𝑑𝑥 − 𝑓(𝑘)| ≤ 2𝜋+1 𝑘2 As a result, 𝑛 𝑘+1 |∑𝑛𝑘=1 𝑓(𝑘) − ∫1 𝑓(𝑥)𝑑𝑥 | ≤ |𝑓(𝑛)| + ∑𝑛−1 𝑘=1 |𝑓(𝑘) − ∫𝑘 𝑓(𝑥)𝑑𝑥 | ≤ 1 + (2𝜋 + 1) ∑∞ 𝑘=1 (2𝜋 + 1)𝜋 2 < ∞. 6 As a result, =1+ 𝑛 𝑛 1 1 lim ∑ 𝑓(𝑘) = lim ∫ 𝑓(𝑥)𝑑𝑥 𝑛→∞ log 𝑛 𝑛→∞ log 𝑛 1 1 𝑘=1 𝐿 = lim ∫0 cos(sin(2𝜋𝑡)) 𝑑𝑡 𝐿 𝐿→∞ Question #4 Given 1 ⋅ 21 + 2 ⋅ 22 + 3 ⋅ 23 + 4 ⋅ 24 + ⋯ + 𝑑 ⋅ 2𝑑 = ∑𝑑𝑟=1 𝑟 ⋅ 2𝑟 how can we infer to the following solution? 2(𝑑 − 1) ⋅ 2𝑑 + 2 Answer: As noted above, observe that ∑𝑑𝑟=1 𝑥 𝑟 = 𝑥(𝑥 𝑑 −1) 𝑥−1 Differentiating both sides and multiplying by x, we find ∑𝑑𝑟=1 𝑟𝑥 𝑟 = 𝑑𝑥 𝑑+2 −𝑥 𝑑+1 (𝑑+1)+𝑥 (𝑥−1)2 Substituting x=2 ∑𝑑𝑟=1 𝑟2𝑟 = 𝑑2𝑑+2 − (𝑑 + 1)2𝑑+1 + 2 = (𝑑 − 1)2𝑑+1 + 2 Question #5 Find the value of 1 sin 2∘ + 2 sin 4∘ + 3 sin 6∘ + ⋯ + 90 sin 1 80∘ Answer: An approach. One may write 𝑑 ∑𝑛𝑘=1 𝑘 sin(𝑘𝑎) = ∑𝑛𝑘=1 (− cos(𝑘𝑎)) 𝑑𝑎 =− 𝑑 𝑑𝑎 ∑𝑛𝑘=0 cos(𝑘𝑎) 1 𝑘2 =− = 𝑑 1 𝑑𝑎 ( + 1 2 𝑎 2 sin 2 sin[(𝑛+ )𝑎] 2 ) (𝑛+1) sin(𝑛𝑎)−𝑛 sin((𝑛+1)𝑎) 4 sin2 𝑎 2 then one may take 𝑎 ≔ 2∘ , 𝑛 ≔ 90 Question #6 How to show that that ∑∞ 𝑛=1 ( 1 3𝑛−1 + 1 3𝑛−2 − 2 3𝑛 ) = ln(3) Answer: Use the fact that 1 1 1 𝑚+1 = ∫0 𝑥 𝑚 𝑑𝑥 to get 3𝑛−2 ) ∑∞ + 𝑥 3𝑛−3 − 2𝑥 3𝑛−1 𝑑𝑥 𝑛=1 ∫0 (𝑥 1 3𝑛−2 = ∫0 ∑∞ + 𝑥 3𝑛−3 − 2𝑥 3𝑛−1 )𝑑𝑥 𝑛=1(𝑥 1 𝑥+1−2𝑥 2 = ∫0 1−𝑥 3 1 2𝑥+1 𝑑𝑥 = ∫0 𝑥 2 +𝑥+1 𝑑𝑥 = log(𝑥 2 + 𝑥 + 1) |10 = log 3 Question #7 I need to find the limit of the following sequence: lim ) 𝑘 𝑛→∞ ∏𝑛 𝑘=1(1+𝑛2 ) Answer: Note that we have log (∏𝑛𝑘=1 (1 + 𝑘 𝑛2 )) = ∑𝑛𝑘=1 log (1 + 𝑘 𝑛2 ) (1) Applying the right-hand side inequality in (1) reveals ∑𝑛𝑘=1 log (1 + = 𝑘 𝑛2 ) ≤ ∑𝑛𝑘=1 𝑘 𝑛2 𝑛(𝑛+1) 1 2𝑛2 = + 2 1 2𝑛 (2) Applying the left-hand side inequality in (1) reveals ∑𝑛𝑘=1 log (1 + ≥ ∑ 𝑘 = 1𝑛 = = 𝑘 𝑛2 𝑘 ) ≥ ∑𝑛𝑘=1 𝑘 𝑘+𝑛2 𝑛+𝑛2 𝑛(𝑛+1) 2(𝑛2 +𝑛) 1 2 (3) Putting 1-3 together yields 1 2 ≤ log (∏𝑛𝑘=1 (1 + 𝑘 𝑛2 1 1 2 2𝑛 )) ≤ + (4) whereby application of the squeeze theorem to (4) gives lim log (∏𝑛𝑘=1 (1 + 𝑛→∞ 𝑘 𝑛2 )) = 1 2 Hence, we find that lim 𝑘 = √𝑒 𝑛→∞ ∏𝑛 𝑘=1(1+𝑛2 ) Question #8 I want to try and evaluate this interesting sum: ∑∞ 𝑛=1 1 Γ(𝑛+𝑠) where 0 ≤ 𝑠 < 𝑞 Answer: We have ∑∞ 𝑛=1 1 = Γ(𝑛+𝑠) 1 Γ(𝑠+1) (1 + 1 1 𝑠+1 + (𝑠+1)(𝑠+2) + ⋯ ) One way to develop such a series is by successively integrating by parts the right hand side of 1 𝑓(𝑠) = ∫0 𝑑𝜁𝑠(1 − 𝜁) 𝑠−1 𝐹(𝜁) 1 where 𝐹(𝜁) is some analytic function of 𝜁 and ∫0 𝐹′ (0) 𝑓(𝑠) = 𝐹(0) + 𝑠+1 is shorthand for lim1−𝜖 . One finds 𝜖→0+ ∫0 𝐹 ′′ (0) + (𝑠+1)(𝑠+2) + ⋯ For this problem we have 𝐹 (𝑛) (0) = 1, so 𝐹(𝜁) = 𝑒 𝜁 . Then 1 ∑∞ 𝑛=1 = = = 𝑓(𝑠) = Γ(𝑛+𝑠) Γ(𝑠+1) 1 1 𝑑𝜁𝑠(1 − 𝜁) 𝑠−1 ∫ Γ(𝑠+1) 0 1 𝑒 ∫ 𝑑𝑢 𝑢 𝑠−1 𝑒 −𝑢 Γ(𝑠) 0 𝑒 Γ(𝑠) 𝑒𝜁 𝛾(𝑠, 1) where 𝛾(𝑠, 𝑥) is the lower incomplete gamma function. Note that 𝛾(𝑠, 𝑥) = Γ(𝑠) − Γ(𝑠, 𝑥) , where Γ(𝑠, 𝑥) is the upper incomplete gamma function. Therefore, ∑∞ 𝑛=1 1 = 𝑒 (1 − Γ(𝑛+𝑠) Γ(𝑠,1) Γ(𝑠) ) as claimed. Question #9 There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows 1 𝑥+1 + 2 𝑥 2 +1 + 4 𝑥 4 +1 +⋯ Answer: By induction, it is pretty clear that 𝑥2 𝑛+1 𝑘 − 1 = (𝑥 − 1) ∏𝑛𝑘=0 (𝑥 2 + 1) 𝑑 and by applying 𝑥 ⋅ 𝑛+1 2𝑛+1 𝑥 2 2𝑛+1 𝑥 −1 = 𝑥 𝑥−1 log(. ) to both sides we have: 𝑑𝑥 + ∑𝑛𝑘=0 2𝑘 𝑥 2 𝑘 𝑥 2 +1 1 and by replacing x with 2𝑛+1 1−𝑧 2 𝑛+1 = 1 1−𝑧 + ∑𝑛𝑘=0 𝑘 𝑧 : 2𝑘 𝑘 𝑧 2 +1 Question #10 Evaluate the integral 𝜋 ∫−𝜋 | ∑∞ 𝑛=1 1 2𝑛 𝑒 ∈𝑥 |2 𝑑𝑥 Answer: Define 𝑔(𝑥) = ∑∞ 𝑛=1 𝑒 𝑖𝑛𝑥 2𝑛 This is absolutely convergent everywhere, and defines a continuous function (by, for example, the Weierstrass M-test). The Fourier coefficients of g are easy to compute: Define 1 𝜋 ∫ 𝑔(𝑥)𝑒 −𝑖𝑛𝑥 𝑑𝑥 2𝜋 −𝜋 By orthogonality of the functions 𝑒 𝑖𝑛𝑥 , we find that 𝑐𝑛 = 𝑐𝑛 = 1 𝜋 𝑒 𝑖(𝑘−𝑛)𝑥 𝑑𝑥 ∫ ∑∞ 2𝜋 −𝜋 𝑘=1 2𝑘 = 1 2𝜋 2𝜋 2𝑛 = 1 2𝑛 Then Parseval's identity shows that 𝜋 2𝜋 2 ∞ −𝑛 = ∫−𝜋|𝑔(𝑥)|2 𝑑𝑥 = 2𝜋 ∑∞ 𝑛=1|𝑐𝑛 | = 2𝜋 ∑𝑛=1 4 3