Questions and Answers #19 Series

Questions and Answers Sheet 19 Series Question #1 Evaluate ∞ ∑ 𝑛=1 𝜙(𝑛) 7𝑛 + 1 Answer: 7 is not special at all. We can replace it with any 𝑎 > 1 (even real numbers). ∞ ∑ 𝑛=1 ∞ ∞ 𝑛=1 𝑛=1 𝜙(𝑛) 𝑎−𝑛 = ∑ 𝜙(𝑛) = ∑ 𝜙(𝑛)𝑎−𝑛 (1 − 𝑎 −𝑛 + 𝑎−2𝑛 − ⋯ ) 𝑛 𝑎 +1 1 + 𝑎 −𝑛 ∞ ∞ = ∑ ∑ 𝜙(𝑛)(−1)𝑘+1 𝑎−𝑘𝑛 𝑛=1 𝑘=1 ∞ = ∑ ∑ 𝑚=1 𝑚 +1 𝑑 𝑑|𝑚(−1)  𝜙(𝑑)𝑎 −𝑚 m d = − a − m ( (−1)  (d )) m =1 d |m Now we will show the following: m is even 0 −m m is odd m  ( −1) d  ( d ) =  d∣ m m  (−1) d  (d ) +  (2d ) + ... +  (d ) = −m . If m is even, let 𝑚 = 2𝑘𝑡 for odd t and If m is odd, then d |m d |t d |t 𝑘 ≥ 1 . Then  (−1) m d d |m  (d ) =  (d ) +  (2d ) + ... +  (2k −1 d ) −  (2k d ) d |t d |t d |t d |t (1 + 𝜙(2) + ⋯ + 𝜙(2𝑘−1 ) − 𝜙(2𝑘 )) ∑𝑑|𝑡𝜙(𝑑)=0 Hence the sum equals to ∞ ∑ 𝑘=0 2𝑘 + 1 𝑎(𝑎2 + 1) = 2 (𝑎 − 1)2 𝑎2𝑘+1 Question #2 How to show that ∑∞ 𝑛=1 1 𝑛2 +3𝑛+1 = 𝜋√5 5 tan 𝜋√5 2 Answer: I think we can make some use of the residue theorem. Write 𝑛2 + 3𝑛 + 1 = (𝑛 + 3/2)2 − 5/4 and the sum ∞ 1 ∞ 1 1 ∑ = ∑ −1+1 2 3 5 2 3 2 5 𝑛=1 (𝑛 + ) − 𝑛=−∞ (𝑛 + ) − 2 4 2 4 (To get the doubly infinite sum, I had to add back the 𝑛 = 0 and 𝑛 = −1 terms, which happen to sum to zero.) The sum on the RHS may be attacked via the residue theorem, using the following: ∞ ∑ 𝑓(𝑛) = −𝜋 ∑ 𝑅𝑒𝑠[𝑓(𝑧) cot 𝜋 𝑧] 𝑛=−∞ 𝑘 where 𝑧𝑘 is a non-integer pole of f. The poles are at 3 √5 2 2 𝑧± = − ± The sum is then equal to 𝜋 2√5 [cot ( 3𝜋 − 2 √5𝜋 2 ) − cot ( 3𝜋 2 + √5𝜋 2 )] = 𝜋 √5 √5𝜋 tan 2 Question #3 (−1)𝑘 Prove ∑∞ 𝑘=1 𝑘 ∑2𝑘 𝑗=1 (−1)𝑗 𝑗 = 𝜋2 48 1 + ln2 2 4 Answer: Use 𝑛 1 𝑥𝑛 (−1) 𝑗−1 ∑ = ln 2 + (−1)𝑛−1 ∫0 1+𝑥𝑑𝑥 𝑗 𝑗=1 we get ∞ 2𝑘 𝑘=1 𝑗=1 ∞ ∞ 𝑘=1 𝑘=1 (−1)𝑘 (−1) 𝑗 (−1)𝑘 (−1)𝑘 1 𝑥 2𝑘 ∑ ∑ = ln 2 ∑ −∑ ∫ 𝑑𝑥 𝑘 𝑗 𝑘 𝑘 0 1+𝑥 1 = ln2 2 + ∫ 0 ∞ 1 ∑ 1+𝑥 𝑘=1 (−𝑥 2 )𝑘 𝑘 1 𝑑𝑥 = ln2 2 − ∫ 0 ln(1 + 𝑥 2 ) 𝑑𝑥 1+𝑥 let 1 ln(1+𝑡𝑥 2 ) 𝑓(𝑡) = ∫0 1+𝑥 𝑑𝑥 then 𝑥2 𝑑𝑥 2 0 (1 + 𝑥)(1 + 𝑡𝑥 ) 1 1 1 𝑥−1 1 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 2 𝑡 + 1 0 1 + 𝑡𝑥 𝑡+1 0 𝑥+1 1 1 1 1 = [ ln(1 + 𝑡𝑥 2 ) − arctan(√𝑡𝑥) + ln(𝑥 + 1)] 𝑡 + 1 2𝑡 √𝑡 0 1 1 1 = [ ln(1 + 𝑡) − arctan(√𝑡) + ln 2] 𝑡 + 1 2𝑡 √𝑡 Integrate back 1 1 ⇒ 𝑓(𝑡) = [−𝐿𝑖2 (−𝑡) − ln2 (𝑡 + 1)] − arctan2√𝑡 + ln 2 ln(𝑡 + 1) 2 2 then let t=1 and use Li_2(-1)=-\frac{\pi^2}{12} , you will get the answer as wanted. 1 𝑓 ′ (𝑡) = ∫ Question #4 Author introduces an alternating series: 𝑛 ∑∞ 𝑛=1(−1) (2𝑛)! 4 𝑛(𝑛!)2 Answer: 𝑛 We are looking at ∑∞ 𝑛=1(−1) If 𝑎𝑛 = (2𝑛)! 4 𝑛(𝑛!)2 (2𝑛)! 4 𝑛(𝑛!)2 (2𝑛)! 𝑎𝑛+1 (2(𝑛 + 1))! = 2 / 𝑛 (𝑛!)2 𝑎𝑛 4𝑛+1 ((𝑛 + 1)!) 4 (2(𝑛 + 1))! 4𝑛 (𝑛!)2 = 2 4𝑛+1 ((𝑛 + 1)!) (2𝑛)! (2𝑛 + 1)(2𝑛 + 2) = 4(𝑛 + 1)2 2𝑛 + 1 = 2(𝑛 + 1) 1 =1− 2(𝑛 + 1) So 𝑎𝑛 is strictly decreasing. Since ∑ converges 𝑎 Also, 𝑛+1 = 1 − 1 𝑎𝑛 𝑎𝑁 𝑎𝑀 = > 1− 2(𝑛+1) 𝑎𝑛+1 𝑛 ∏𝑁−1 ∏𝑁−1 < 𝑛=𝑀 𝑎 𝑗=𝑀 𝑛+1 𝑛 1 𝑛+1 𝑀 = 1 2(𝑛+1) 𝑛 𝑛+1 diverges 𝑎𝑛 → 0 . By the alternating series test, ∑(−1)𝑛 𝑎𝑛 so 1 = . Therefore, since ∑ diverges, ∑ 𝑎𝑛 𝑁 𝑁 Question #5 I was able to prove this sum ∑∞ 𝑛=1 1 𝑛+3𝑛 Answer: What we can obtain, also by hand, is a reasonable estimation for example by 1 1 1 ∑∞ ∑3𝑛=1 ∑∞ 𝑛=4 𝑛 = 𝑛=1 𝑛 ≈ 0.392 < 𝑛 + 1 𝑛+3 1 1 4 11 30 = + + 3 1 𝑛+3 1 1 2 3 9 + −1− − − 27 3 ≈ 0.393 Question #6 ∑∞ 𝑛=1 (sin 𝑛+2)𝑛 𝑛3𝑛 Does it converge or diverge? Answer: The values for which sin(𝑛) is close to 1 (say in an interval [1 − 𝜖; 1] ) are some what regular: 𝜋 1 − 𝜖 ≤ sin(𝑛) implies that there exists an integer k(n) such that 𝑛 = 2𝑘(𝑛) + 𝜋 + + 𝑎(𝑛) where 2 |𝑎(𝑛)| ≤ arccos(1 − 𝜖). As 𝜖 → 0, arccos(1 − 𝜖) ∼ √2𝜖, thus we can safely say that for 𝜖 small enough, | n − 2k (n) −  2 |=| a (n) | 2 If m>n and since sin(𝑛) and sin(𝑚) are both in [1 − 𝜖; 1], then we have the inequality | ( m − n) − 2( k ( m) − k ( n)) || m − 2k ( m) −  2 | + | n − 2k ( n) −  2 | 4 where (k(m)-k(n)) is some integer k. Since 𝜋 has a finite irrationality measure, we know that there is a finite real constant 𝜇 > 2 such that for any integers n,k large enough, |𝑛 − 𝑘𝜋| ≥ 𝑘1−𝜇 By picking 𝜖 small enough we can forget about the finite number of exceptions to the inequality, and we get 4√𝜖 ≥ (2𝑘)1−𝜇 . Thus 1 1 (𝑚 − 𝑛) ≥ 2𝑘𝜋 − 4√𝜖 ≥ 𝜋(4√𝜖)1−𝜇 − 4𝜖 ≥ 𝐴𝜖 = 𝐴√𝜖 1−𝜇 for some constant A. Therefore, we have a guarantee on the lengh of the gaps between equally problematic terms, and we know how this length grows as \epsilon gets smaller We can get a lower bound for the first problematic term using the irrationality measure as well : from | n − 2k (n) −  2 | 2 , we get that for 𝜖 small enough, (4𝑘 + 1)1−𝜇 ≤ |2𝑛 − (4𝑘 + 1)𝜋| ≤ 4√𝜖, and 1 then 𝑛 ≥ 𝐵𝜖 = 𝐵 √𝜖 1−𝜇 for some constant B. Therefore, there exists a constant C such that forall 𝜖 small enough, the k-th integer n such that 1 − 𝜖 ≤ 1 sin 𝑛 is greater than 𝐶𝜖 𝑘 = 𝐶 √𝜖 1−𝜇 𝑘 1 Since 𝜖 < 1 and 1−𝜇 < 0 , this bound 𝐶𝜖 grows when 𝜖 gets smaller. And furthermore, the speed of this growth is greater if we can pick a smaller (better) value for 𝜇 (though all that matters is that 𝜇 is finite) Now let us give an upper bound on the contribution of the terms where n is an integer such that sin(𝑛) ∈ [1 − 2𝜖; 1 − 𝜖] 𝑆𝜖 = ∑ =  (1+sin(𝑛))𝑛 𝑛3𝑛 ≤ ∑𝑘≥1 (1−𝜖/3)𝑘𝐶2𝑘 𝑘𝐶2∈ − log(1−(1−𝜖/3)𝐶2𝑥 ) 𝐶2𝜖 − log(1 − (1 − C2 / 3)) − log(C2 / 3)) = C2 C2 1 𝐶2𝜖 = 𝐶 √2𝜖 1−𝜇 = 𝐶 ′ 𝜖 𝑣 with 𝑣 = 𝑆𝜖 ≤ − 1 2(1−𝜇) ∈ [−1/2; 0], so: log(𝐶 ′ /3)+(1+𝑣) log 𝜖 𝐶 ′ 𝜖𝑣 Finally, we have to check if the series ∑ 𝑆2−𝑘 converges or not : ∑ 𝑆2−𝑘 ≤ ∑ − log(𝐶 ′ /3)−𝑘(1+𝑣) log 2 𝐶 ′2−𝑘𝑣 = ∑(𝐴 + 𝐵𝑘)(2𝑣 )𝑘 Question #7 Prove: 2𝜋 4𝜋 6𝜋 8𝜋 5 5 5 5 cos ( ) + cos ( ) + cos ( ) + cos ( ) = −1 Answer: If 𝑧 5 − 1 = 0, then (𝑧 4 + 𝑧 3 + 𝑧 2 + 𝑧 + 1)(𝑧 − 1) = 0 because 𝑧 5 − 1 = (𝑧 4 + 𝑧 3 + 𝑧 2 + 𝑧 + 1)(𝑧 − 1). This implies that 𝑧4 + 𝑧3 + 𝑧2 + 𝑧 + 1 = 0 or 𝑧 − 1 = 0 . Since 2𝜋 2𝜋 𝑧 = cos + 𝑖 sin = 𝑒 𝑖2𝜋/5 ≠ 1 5 5 is a root of 𝑧 5 − 1 = 0 and 2𝑘𝜋 2𝑘𝜋 𝑧 𝑘 = cos + 𝑖 sin = 𝑒 𝑖2𝑘𝜋/5 , 5 5 for 𝑘 ∈ [1,2,3,4] we have 8𝜋 8𝜋 6𝜋 6𝜋 0 = 𝑧 4 + 𝑧 3 + 𝑧 2 + 𝑧 + 1 = (cos + 𝑖 sin ) + (cos + 𝑖 sin )+ ⋯+ 1 5 5 5 5 8𝜋 6𝜋 8𝜋 6𝜋 2𝜋 = (cos + cos + ⋯ + 1) + 𝑖 (sin + cos + ⋯ + sin ) So cos 5 8𝜋 5 + cos 5 6𝜋 5 5 + ⋯+ 1 = 0 Question #8 Find the sum of series: ∑∞ 𝑛=0 (4𝑛)! (4𝑛+4)! 5 5 Answer: (4𝑛)! Γ(4𝑛+1) = ∑𝑛≥0 (4𝑛+4)! Γ(4𝑛+5) 1 ∑𝑛≥0 𝐵(4𝑛 + 1,4) 6 1 1 ∑𝑛≥0 ∫0 𝑥 4𝑛 (1 − 𝑥)3 𝑑𝑥 6 (1−𝑥)2 1 1 𝑑𝑥 ∫ 6 0 (1+𝑥 2 )(1+𝑥) ln 2 𝜋 ∑𝑛≥0 = = = = − 4 24 Question #9 Sum of the series: 1+𝑛 ∑∞ 𝑛=0 3𝑛 Answer: Using the usual geometric series, you have 𝑥 𝑛 = ∑∞ 𝑛=1 𝑥 1−𝑥 thus differentiating you get 1 𝑛−1 ∑∞ = 𝑛=1 𝑛𝑥 2 (1−𝑥) thus 𝑥 𝑛 ∑∞ 𝑛=1 𝑛𝑥 = (1−𝑥)2 𝑥 𝑛 Then ∑∞ 𝑛=1 𝑛𝑥 = (1−𝑥)2 So your sum is just ∞ ∑ 𝑛=1 𝑛 1/3 3 = = 𝑛 2 (1 − 1/3) 3 4 Question #10 Proof: 𝑛2 +3𝑛+2 ∑∞ 𝑛=0 4𝑛 = 128 27 Given hint: (𝑛2 + 3𝑛 + 2) = (𝑛 + 2)(𝑛 + 1) Answer: Using the formula for the sum of a geometric series: 1 𝑘 ∑∞ (1) 𝑘=0 𝑥 = 1−𝑥 Taking the derivative of (1) and tossing the term which are 0: 1 𝑘−1 ∑∞ = (1−𝑥)2(2) 𝑘=1 𝑘𝑥 Taking the derivative of (2) and tossing the terms which are 0: 2 𝑘−2 ∑∞ = (1−𝑥)3 (3) 𝑘=2 𝑘(𝑘 − 1)𝑥 Reindexing the sum in (3): 2 𝑘 ∑∞ 𝑘=0(𝑘 + 2)(𝑘 + 1)𝑥 = (1−𝑥)3 1 Plug in 𝑥 = : ∑∞ 𝑘=0 = 4 (𝑘+2)(𝑘+1) 128 27 4𝑘 = 2 3 3 4 ( )