Questions and Answers #8 Series

Questions and Answers Sheet 8 Series Question #1 How can I evaluate 2𝑛 ∑∞ 𝑛=1 3𝑛+1 Answer: No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: 𝑛 𝑆𝑚 = ∑𝑚 𝑛=1 𝑛𝑟 Notice that 𝑛 𝑆𝑚 − 𝑟𝑆𝑚 = −𝑚𝑟 𝑚+1 + ∑𝑚 𝑛=1 𝑟 = −𝑚𝑟 𝑚+1 + = 𝑟−𝑟 𝑚+1 1−𝑟 𝑚𝑟 𝑚+2 −(𝑚+1)𝑟 𝑚+1 +𝑟 1−𝑟 Hence 𝑆𝑚 = 𝑚𝑟 𝑚+2 −(𝑚+1)𝑟 𝑚+1 +𝑟 1−𝑟 Hence 𝑆𝑚 = 𝑚𝑟 𝑚+2 −(𝑚+1)𝑟 𝑚+1 𝑟 (1−𝑟)2 1 2 3 3 This equality holds for any r, but in your case we have 𝑟 = and a factor of 1 𝑚( ) 2𝑛 2 3 ∑ 𝑛+1 = lim 3 3 𝑚→∞ ∞ 𝑚+2 1 − (𝑚 + 1) ( ) 3 1 (1 − ( )) 3 𝑛=1 𝑚+1 in front of the sum. That is 1 +( ) 3 2 1 = 2 ( 3) 3 (2)2 3 = 1 2 Question #2 What's the limit of the sequence lim 𝑛! 𝑛→∞ 𝑛𝑛 Answer: You can prove that 𝑛! < ( 𝑛+1 𝑛 2 ) Now observe that 0 ≤ lim 𝑛! 𝑛→∞ 𝑛𝑛 < lim 𝑛→∞ ( 𝑛+1 𝑛 ) 2 𝑛𝑛 = lim 1 𝑛→∞ 2𝑛 ⋅ (𝑛+1)𝑛 𝑛𝑛 We know that 1/2𝑛 → 0 as 𝑛 → ∞. If you know that [(𝑛 + 1)/𝑛]𝑛 → 𝑒 as 𝑛 → ∞, then you're done... the limit is zero! Definitely not as nice as anon's solution, but a different approach nontheless. Question #3 What's the value of ∑∞ 𝑘=1 𝑘2 ? 𝑘! Answer: 𝑘𝑛 The value of 𝑇𝑛 ≔ ∑∞ 𝑘=1 𝑘! 𝑖𝑠𝐵𝑛 ⋅ 𝑒 , where 𝐵𝑛 is the n-th Bell number. To see this, note that 𝑇𝑛+1 = ∑∞ 𝑘=1 𝑘 𝑛+1 1 𝑘! = ∑∞ 𝑘=0 (𝑘+1)𝑛 𝑘! 𝑛 𝑗 = ∑∞ 𝑘=0 ∑𝑗=0((𝑛), (𝑗))𝑘 𝑘! = 𝑗 𝑘 ∑𝑛𝑗=0((𝑛), (𝑗)) ∑∞ 𝑘=1 𝑘! ∑𝑛𝑗=0((𝑛), (𝑗))𝑇𝑗 = This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by e. Question #4 Showing that: 𝑛 √𝑛! lim 𝑛 𝑛→∞ = 1 𝑒 Answer: Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach. Maybe rewrite the left-hand side as 𝑛! √ 𝑛 (𝑛 𝑛 ) Take the logarithm. We get 1 𝑛 1 2 3 𝑛 𝑛 𝑛 𝑛 𝑛 (log ( ) + log ( ) + log ( ) + ⋯ + log ( )) Now think of the above sum as a Riemann sum for the not quite proper integral 1 ∫0 log 𝑥 𝑑𝑥 Question #5 How can we prove that? ∞ ∑ 𝑛=1 3𝑥 𝑥 3𝑛 1 −𝑥 𝜋 + 3√3𝑥 = 𝑒 2 𝑥 (𝑒 2 − 2 sin ( )) (3𝑛 − 1)! 3 6 Answer: Consider the third root of unity 𝜌 = 𝑒 2𝜋𝑖/3 = 𝑒 𝜌𝑧 = ∑∞ 𝑘=0 since 𝜌 3𝑚 𝜌𝑘 𝑧 𝑘 𝑘! 𝑧 3𝑚 2 ∞ = ∑∞ 𝑚=0 (3𝑚)! + 𝜌 ∑𝑚=0 (3𝑚+1)! + 𝜌 . You have 𝑧3𝑚+2 2 ∑∞ 𝑚=0(3𝑚+2)! 2 = 1, 𝜌3𝑚+1 = 𝜌, 𝜌3𝑚+2 = 𝜌2 . You have something similar for 𝑒 𝜌 𝑧 . Also consider 1 + 𝜌 + 𝜌2 = 0. Then a suitable combination of 𝑒 𝜌 ∑∞ 𝑛=1 𝑧 3𝑚+1 −1+𝑖√3 𝑘𝑥 gives you 𝑥 3𝑛−1 (3𝑛−1)! Using Euler's formula 𝑒 𝑖𝑡 = cos 𝑡 + 𝑖 sin 𝑡 then gives you the right hand side. Question #6 How to compute ∏∞ 𝑛=2 𝑛3 −1 𝑛3 +1 Answer: HINT: If 𝑡𝑛 = 𝑛3 −1 = 𝑛3 +1 (𝑛−1)(𝑛2 +𝑛+1) (𝑛+1)(𝑛2 −𝑛+1) 𝑛[(𝑛+1)2 +𝑛+1+1] 𝑡𝑛+1 = (𝑛+2)[(𝑛+1)2 −(𝑛+1)−1] = 𝑛[(𝑛+1)2 +𝑛+1+1] (𝑛+2)(𝑛2 +𝑛+1) and 𝑡𝑛−1 = (𝑛−2)[(𝑛−1)2 +𝑛−1+1] 𝑛[(𝑛−1)2 −(𝑛−1)−1] Alternatively, let 𝑢𝑛 = ⇒ ∏2≤𝑛≤𝑟 𝑢𝑛 = = (𝑛−2)(𝑛2 −𝑛+1) 𝑛[(𝑛−1)2 −(𝑛−1)−1] 𝑛2 +𝑛+1 ,𝑣 = 2 so that 𝑛+1 𝑛 𝑛 −𝑛+1 𝑛−1 1⋅2⋅3⋅…(𝑟−2)(𝑟−1) 2 3⋅4⋅5⋅…𝑟(𝑟+1) 𝑟(𝑟+1) ⇒ ∏2≤𝑛≤𝑟 𝑡𝑛 = (∏2≤𝑛≤𝑟 𝑢𝑛 )(∏2≤𝑛≤𝑟 𝑣𝑛 ) = Setting 𝑟 → ∞, 2(𝑟 2 +𝑟+1) 3𝑟(𝑟+1) 1 2 = lim 𝑡𝑛 = 𝑢𝑛 ⋅ 𝑣𝑛 2(𝑟 2 +𝑟+1) 3𝑟(𝑟+1) 1 1+𝑟+ 2 𝑟 3 𝑟→∞ 1+ 1 𝑟 Question #7 Evaluate the infinite product lim 1 𝑛→∞ ∏𝑛 𝑘=2(1−𝑘2 ) ) Answer: Note that 1 1 1 𝑘 − 1𝑘 + 1 𝑎𝑘 1 − 2 = (1 − ) (1 + ) = = 𝑘 𝑘 𝑘 𝑘 𝑘 𝑎𝑘−1 with 𝑎𝑘 = 𝑛 ∏ (1 − 𝑘=2 𝑘+1 𝑘 , hence this is a telescoping product, i.e. 1 𝑎2 𝑎3 𝑎𝑛 𝑎𝑛 𝑛 + 1 )= … = = 𝑘2 𝑎1 𝑎2 𝑎𝑛−1 𝑎1 2𝑛 Question #8 What is the closed form for ∑∞ 𝑛=−∞ ? Answer: In my opinion, this is a rather nice example of application of the Poisson summation formula: ∑𝑛∈𝑍 𝑓(𝑛) = ∑𝑘∈𝑍 𝑓̂ (𝑘) , ∞ 𝑓̂(𝑣) = ∫−∞ 𝑓(𝑥)𝑒 −2𝜋𝑖𝑣𝑥 𝑑𝑥 1 Namely, setting 𝑓(𝑥) = (𝑥−𝑎)2 +𝑏 2 , we find ∞ 𝑒 −2𝜋𝑖𝑣𝑥𝑑𝑥 𝑒 −2𝜋𝑖𝑣𝑥 𝑑𝑥 𝜋 −2𝜋𝑖𝑣𝑎−2𝜋|𝑣|𝑏 −2𝜋𝑖𝑣𝑎 ∫−∞ 𝑥 2 +𝑏 2 = = 𝑒 𝑒 2 + 𝑏2 (𝑥 − 𝑎) 𝑏 −∞ where we have assumed that b>0 and calculated the last integral using residues. Therefore, the sum we are trying to calculate reduces to geometric series: 𝑓̂(𝑣) = ∫ ∞ ∑ 𝑛∈𝑍 1 𝜋 𝜋 sinh 2 𝜋𝑏 = ∑ 𝑒 −2𝜋𝑖𝑘𝑎−2𝜋|𝑘|𝑏 = (𝑛 − 𝑎)2 + 𝑏 2 𝑏 𝑏(cosh 2 𝜋𝑏 − cos 2 𝜋𝑎) 𝑘∈𝑍 Question #9 𝜋3 Proving that = 1 − ∑∞ 𝑘=1 32 2𝑘(2𝑘+1)𝜁(2𝑘+2) 4 2𝑘+2 Answer: Yes, we can prove it. We can change the order of summation in ∑∞ 𝑘=1 2𝑘(2𝑘+1)𝜁(2𝑘+2) ∞ ∑∞ 𝑛=1 ∑𝑘=1 4 2𝑘+2 2𝑘(2𝑘+1) = ∑∞ 𝑘=1 2𝑘(2𝑘+1) 4 2𝑘+2 ∑∞ 𝑛=1 1 𝑛2𝑘+2 (4𝑛)2𝑘+2 ∞ ′′ ∑𝑛=1 𝑟 (4𝑛) where, for |z|>1 , we define 1 𝑟(𝑧) = ∑∞ 𝑘=1 𝑧 2𝑘 1 = 1 = ( 𝑧 2 −1 Differentiating yields 𝑟 1 − ∑∞ 𝑘=1 1 2 𝑧−1 ′′ (𝑧) 2𝑘(2𝑘+1)𝜁(2𝑘+2) 4 𝑘+2 − 1 𝑧+1 1 ) 1 = (𝑧−1)3 − (𝑧+1)3, so (−1)𝑣 = ∑∞ 𝑣=0 (2𝑣+1)3 Using the partial fraction decomposition of ∑∞ 𝑣=0 (−1)𝑣 (2𝑣+1)3 =− 𝜋3 32 𝐸2 = 𝜋3 1 cos 𝑧 : 32 Question #10 Prove that: ∑∞ 𝑘=0 (−1)𝑘 (2𝑘+1)3 = 𝜋3 32 Answer: You can evaluate this sum using the residue theorem. First, note that it may be extended out to −∞ : ∞ ∞ 𝑘=0 𝑘=−∞ (−1)𝑘 (−1)𝑘 1 ∑ = ∑ (2𝑘 + 1)3 2 (2𝑘 + 1)3 From the residue theorem: ∑∞ 𝑘=−∞ (−1)𝑘 (2𝑘+1)3 𝜋 csc 𝜋𝑧 1 𝜋 csc 𝜋𝑧 = −𝑅𝑒𝑠𝑧=−1/2 (2𝑧+1)3 = − 𝑅𝑒𝑠𝑧=−1/2 (𝑧+1/2)3 8 This residue involves taking the second derivative of the csc term. Note that, for a generic function f(z) having a triple pole at 𝑧 = 𝑧0 : 𝑅𝑒𝑠𝑧=𝑧0 𝑓(𝑧) = 1 lim 𝑑2 2! 𝑧→𝑧0 𝑑𝑧 2 [(𝑧 − 𝑧0 )3 𝑓(𝑧)] so that 𝜋 csc 𝜋𝑧 𝑅𝑒𝑠𝑧=−1/2 (𝑧+1/2)3 = 𝜋3 2! [csc 𝜋 𝑧 cot 2 𝜋𝑧 + csc 3 𝜋𝑧]𝑧=−1/2 = − Putting this all together, we get the stated result: ∑∞ 𝑘=0 (−1)𝑘 (2𝑘+1)3 = 𝜋3 32 𝜋3 2