Questions and Answers #8 Sequences

Questions and Answers Sheet 8 Sequences Question #1 Find 𝑎5 and 𝑎𝑛 for the geometric sequences. 𝑎2 = 12, 𝑟 = 1/2 Answer: Common ratio: 𝑟 = 1/2 General term of the geometric sequence 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 Second term 𝑎2 = 12 Thus, we have 𝑎2 = 12 𝑎1 𝑟 2−1 = 12 1 1 𝑎1 ( ) = 12 2 𝑎1 = 24 1 𝑛−1 Thus, the general term is 𝑎𝑛 = 24 ( ) 2 Find 𝑎5 𝑎5 = 𝑎1 𝑟 5−1 1 4 = 24 ( ) 2 1 = 24 ( ) 16 = 3 2 Question #2 Find the false statement 1) ∑∞ 𝑛=2 1 𝑛(ln 𝑛)𝑃 converges if p>1 2) the integral test does not apply to divergent sequence 3) ∑∞ 𝑛=1 1 𝑛𝑃 converges if p>1 and diverges if 𝑝 ≤ 1 ∞ 4) If 𝑎𝑛 and f(n) satisfy the requirements of the integral test, and if ∫1 𝑓(𝑥)𝑑𝑥 converges, then ∞ ∑∞ 𝑛=1 𝑎𝑛 = ∫1 𝑓(𝑥)𝑑𝑥 Answer: Consider the statements: (a) ∑∞ 𝑛=2 1 𝑛(ln 𝑛)𝑃 converges if p>1 by Cauchy condensation test Hence, (a) option is true. (b) The integral test does not apply to divergent sequences The given statement is false. 1 ∑∞ 𝑛=1 𝑎𝑛 𝑏𝑛 < that is convergent and if 𝑎𝑛 → 0 then | an |  Thus, the second statement is false. Question #3 Find the following sums of arithmetic sequences. 881+872+863+854+....8 Answer: 881+872+863+854+....8 Gauss Method: Sum of terms=Average term\times Number of terms Average\ term = = = 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚+𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 2 881+8 2 889 2 -Average term Number of terms: 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 − 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 ( )+1 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 − 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑒𝑟𝑚 = 881 − 8 +1 881 − 872 = 873 +1 9 = 97 + 1 =98 - Number of terms Therefore, the Sum of terms 889 × 98 = 43561 2 The answer: 43561 Question #4 Find the limit 𝑎𝑛 = 8 − 5𝑛 7𝑛 Answer: Given: 𝑎𝑛 = 8−5𝑛 7𝑛 then, lim 𝑎𝑛 = lim 𝑛→∞ 8−5𝑛 𝑛→∞ 7𝑛 Divide by n 8 −5 𝑛 = lim ( ) 𝑛→∞ 7 We know that lim 1 𝑛→∞ 𝑛 = 0−5 7 = −5 7 = 0, so Question #5 Find the limit of the sequence 𝑎𝑛 = 5𝑛2 − 8𝑛 + 11 7𝑛 + 4 Answer: Given: 𝑎𝑛 = lim 𝑛→∞ 5𝑛2 − 8𝑛 + 11 7𝑛 + 4 5𝑛2 −8𝑛+11 7𝑛+4 Divide numerator and denominator by 𝑛2 5𝑛2 − 8𝑛 + 11 𝑛2 = lim 7𝑛 + 4 𝑛→∞ 𝑛2 8 11 + 𝑛 𝑛2 7 4 + 𝑛 𝑛2 5− = lim 𝑛→∞ = 5−0+0 0+0 =∞ Question #6 Find the n-th term -5, -3, -1, 1, ... Answer: Given : -5, -3, -1, 1, ... 𝑎1 = −5 𝑑 = 𝑎2 − 𝑎1 = −3 − (−5) = −3 + 5 = 2 𝑑 = 𝑎3 − 𝑎2 = −1 − (−3) = −1 + 3 = 2 This is an arithmetic sequence, thus, general form is 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 Question #7 Find the seventh term of Fibonacci sequences: (1) 5, 10, 30, 120,... , (2) 2, 1, 1/3, 1/12,... Answer: 1) 5, 10, 30, 120,... 5 × 2 = 10 10 × 3 = 30 30 × 4 = 120 Hence, 5-th term is 120 × 5 = 600 6-th term is 600 × 6 = 3600 7-th term is 3600 × 7 = 25200 (answer) Question #8 Find the sum of the infinite geometric sequences: 48, 24, 12... Answer: Given: 48, 24, 12... First term a=48 Common ratio 𝑟 = 24 48 = 1 2 Sum of inifinte terms 𝑆∞ = 𝑎 1−𝑟 48 𝑆∞ = 1− 1 2 48 1 2 𝑆∞ = 𝑆∞ = 48 × 2 𝑆∞ = 96 Question #9 Find the sum of the terms from 𝑎16 to 𝑎23 inclusive. Use the sum formula for arithmetic sequences. 7, 10, 13, 16,... Answer: Given 7, 10, 13, 16,... First term a=7 Common difference d=20-7=3 We need to find the sum of terms from 𝑎16 to 𝑎23 Sum of n terms: 𝑆𝑛 = 𝑛 (2𝑎 + (𝑛 − 1)𝑑) 2 Sum of first 15 terms 𝑆15 = 15 (2(7) + (15 − 1)(3)) 2 = 15 [14 + 14(3)] 2 = 15 [56] 2 = 420 Sum of first 23 terms 𝑆23 = = 23 (2(7) + (23 − 1)(3)) 2 23 [14 + 66] 2 = 920 The sum of terms from 𝑎16 to 𝑎23 = 𝑆23 − 𝑆15 = 920 − 420 = 500 Question #10 Find the sum of the terms from 𝑎16 to 𝑎23 for the following sequence 7, 10, 13, 16, ... Answer: The general term 𝑎𝑛 of 𝑎𝑛 arithmetic sequence is defined using the formula: 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 The first term a=7 The common difference d=10-7=3 Substitute a=7, d=3 in the formula 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 7 + (𝑛 − 1)3 = 7 + 3𝑛 − 3 = 3𝑛 + 4 Hence, 𝑎𝑛 = 3𝑛 + 4 Calculate 𝑎16 by substituting n=16 in the general term. 𝑎𝑛 = 3𝑛 + 4 𝑎16 = 3(16) + 4 = 48 + 4 = 52 Calculate 𝑎23 by substituting n=23 in the general term. 𝑎𝑛 = 3𝑛 + 4 𝑎23 = 3(23) + 4 = 69 + 4 = 73 The total number of terms between 𝑎16 and 𝑎23 are 8. Let it be an arithmetic sequence, whose first term is 𝑎16 and the last term is 𝑎23 . The sum of these terms 𝑆𝑛 : 𝑆𝑛 = 𝑛 [𝐹𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 + 𝐿𝑎𝑠𝑡 𝑡𝑒𝑟𝑚] 2 8 𝑆8 = [𝑎16 + 𝑎23 ] 2 8 = [52 + 73] 2 = 4 × 125 = 500 Question #11 Find the value of 75th term in this sequences: 5/2, 5, 10, 20, 40, 80, ... Answer: Given sequence 5 , 5,10,20,40,80, … 2 Now 5 5 2 = 2, 10 5 = 2, 20 10 =2 Thus, the sequence is in geometric progression First term 𝑎 = 5 2 and common ratio r=2 The formula for n-th term: 𝑎𝑛 = 𝑎𝑟 𝑛−1 ∴ 𝑎75 = 𝑎75 = 5 × 275−1 2 5 × 274 2 𝑎75 = 5 × 273 Thus, the answer is 5 × 273 Question #12 Find the z-transform f(z) 𝑓(𝑛) = (0.5)𝑛 𝑢(𝑛) + (1.5)𝑛 𝑢(−𝑛 − 1) − 2δ(𝑛) Answer: Given 𝑓(𝑛) = (0.5)𝑛 𝑢(𝑛) + (1.5)𝑛 𝑢(−𝑛 − 1) − 2δ(𝑛) In z-transform 𝑥 𝑛 𝑢(𝑛) ↔ 1 2 (𝑜𝑟) , |𝑧| > 𝑥 1 − 𝑥𝑧 −1 2−𝑎 𝑥 𝑛 𝑢(−𝑛 − 1) ↔ − 𝐹(𝑧) = 1 1−𝑥𝑧 −1 (𝑜𝑟) −𝑧 𝑧−𝑎 , |𝑧| < 𝑥 𝑧 𝑧 − −2 𝑧 − 0.5 𝑧 − 1.5 0.5<|z|<1.5 Question #13 1 𝑛 Find the Z-transform for the sequence 𝑥[𝑛] = 2𝑛 𝑢[−𝑛] + ( ) 𝑢[𝑛 − 1] 4 Answer: 1 𝑛 Given: 𝑥[𝑛] = 2𝑛 𝑢[−𝑛] + ( ) 𝑢[𝑛 − 1] 4 1 𝑛+1−1 Now, 𝑥[𝑛] = (2−1 )−𝑛 𝑢[−𝑛] + ( ) 4 𝑢[𝑛 − 1] 1 −𝑛 1 𝑛−1 1 𝑥[𝑛] = ( ) 𝑢[−𝑛] + ( ) ( ) 𝑢[𝑛 − 1] 2 4 4 1 −𝑛 1 1 𝑛−1 𝑥[𝑛] = ( ) 𝑢[−𝑛] + [( ) 𝑢[𝑛 − 1]] 2 4 4 Apply z-transform 𝑥[𝑧] = [ 𝑥[𝑧] = = 𝑧 1 𝑧− 2 1 𝑧 1 1 − 𝑧 4 1 𝑧 ]+ ( × 𝑧 −1 ) 4 𝑧−1 4 1 1 + ( ) 4 𝑧−1 4 1 2𝑧 1 1 × + ( ) 𝑧 2−𝑧 4 𝑧−1 4 𝑥[𝑧] = −2 1 1 + ( ) 𝑧−2 4 𝑧−1 4 𝑥[𝑧] = 1 1 2 ( )−( ) 4 𝑧−1 𝑧−2 4 1 1 (𝑧 − 2) − 2 (𝑧 − ) 4 4 ] =[ 1 (𝑧 − 2) (𝑧 − ) 4 z 1 2 − − 2z + ) 4 x[ z ] = 4 2 1 ( z − 2)( z − ) 4 𝑥[𝑧] = 𝑥[𝑧] = 7 − 𝑧+0 4 1 (𝑧 − 2) (𝑧 − ) 4 7 4 (− )𝑧 1 4 (𝑧−2)(𝑧− ) The zero of x[z] is the numerator value 7 − 𝑧=0 4 𝑧=0 The pole of x[z] is the determinator value 1 (𝑧 − 2) (𝑧 − ) = 0 4 𝑧 = 2, 1 4 The Radius of convergence 1 𝑛 𝑥[𝑛] = 2𝑛 𝑢[−𝑛] + ( ) 𝑢[𝑛 − 1] 4 1 −𝑛 1 1 𝑛−1 = ( ) 𝑢[−𝑛] + [( ) 𝑢[𝑛 − 1]] 2 4 4 1 −𝑛 1 1 𝑛−1 𝑥[𝑛] = ( ) 𝑢[−𝑛] + [( ) 𝑢[𝑛 − 1]] 2 4 4 𝑥[𝑛] = (1) + (2) Since 𝑎𝑛 𝑥(𝑛) ⇒ |𝑧| > 𝑎, 𝑢[𝑛] ⇒ |𝑧| > 1 𝑎−𝑛 𝑥(−𝑛) ⇒ |𝑧| < ROC (1) ⇒ 𝑎 = |𝑧| < 1 1 2 |𝑧| < 2 1 2 1 𝑛 , 𝑎 𝑢[𝑛] ⇒ |𝑧| > 𝑎 𝑎 ROC (2) ⇒ 𝑎 = |𝑧| > 1 4 1 4 Radius of convergence is the intersection of ROC (1) and ROC (2) 1 < |𝑧| < 2 4 0.25 < |𝑧| < 2 Question #14 Find the z-transform for the sequences 1) x(n)=10u(n) 2) 𝑥(𝑛) = 10 sin(0.25π𝑛) 𝑢(𝑛) Answer: 1) x(n)=10u(n) From the table of z-transformation: 𝑎𝑢(𝑛) = 𝑎𝑧 𝑧−1 Thus, 𝑥(𝑧) = 𝑧(10𝑢(𝑛)) 𝑥(𝑧) = 10𝑧 𝑧−1 2) 𝑥(𝑛) = 10 sin(0.25π𝑛) 𝑢(𝑛) From the table of z-transformation sin(𝑎𝑛 ) 𝑢(𝑛) = 𝑧2 𝑧 sin(𝑎) − 2𝑧 cos 𝑎 + 1 Thus, 𝑥(𝑧) = 10𝑧(sin(0.2π𝑛))𝑢(𝑛) = = 𝑧2 10 sin(0.25π) 𝑧 − 2𝑧 cos(0.25π) + 1 𝑧2 7.007𝑧 − 1.414 + 1 Question #15 Find z-transform of the sequence 1 𝑛 𝑥[𝑛] = ( ) 𝑢[𝑛 + 3] 2 Answer: 1 𝑛 Given: 𝑥[𝑛] = ( ) 𝑢[𝑛 + 3] 2 ∞ 1 𝑛 𝑋(𝑧) = ∑ ( ) 𝑧 −𝑛 2 𝑛=−3 ∞ 𝑛 1 = ∑ [ 𝑧 −1 ] 2 𝑛=−3 𝑁2 ∑ 𝑎𝑛 = 𝑛=𝑁1 𝑎𝑁1 − 𝑎𝑁2 +1 1−𝑎 −3 ∞+1 1 1 ( 𝑧 −1 ) − ( 𝑧 −1 ) 2 𝑋(𝑧) = 2 1 −1 1−( 𝑧 ) 2 −3 1 ( 𝑧 −1 ) 𝑧 3 − 0 2 = 1 1 − 𝑧 −1 2 = 8𝑧 3 × 𝑧 1 𝑧− 2 = 8𝑧 4 1 𝑧− 2 Poles: 𝑧− 1 =0 2 𝑧 = 0.5 Zeros: 8𝑧 4 = 0 𝑧=0 Thus, 1 −1 z | 1 2 1 | z | 2 | Question #16 If P=8+10+12+14+...+108 and Q=4+6+8+10...+106 are sums of arithmetic sequences, determine which of them is greater. Answer: For P=8+10+12+14+...+108 a=8 , d=10-8=2 Let n be total number of terms 108 = 𝑎 + (𝑛 − 1)𝑑 108 = 8 + (𝑛 − 1)2 (𝑛 − 1)2 = 100 𝑛 − 1 = 50 𝑛 = 51 Thus, the sum is: 𝑆𝑃 = 51 (8 + 108) = 2958 2 For Q=4+6+8+10...+106 a=4 , d=6-4=2 Let n be total number of terms 106 = 𝑎 + (𝑛 − 1)𝑑 106 = 4 + (𝑛 − 1)2 (𝑛 − 1)2 = 102 𝑛 − 1 = 51 𝑛 = 52 𝑆𝑄 = 52 (4 + 106) = 2860 2 The difference of sums: 2958-2860=98 Question #17 Let G be a group and let F be the family of all subgroups of finite index. Let 𝑥𝑛 be a sequence in G. We define this sequence to be Cauchy if given H\in F there exists n_0 such that for 𝑚, 𝑛 ≥ 𝑛0 we have −1 𝑥𝑛 𝑥𝑚 ∈ 𝐻 . We define multiplication of two sequences componentwise. Now, we define 𝑥𝑛 to be a null sequence if given 𝐻 ∈ 𝐹 there exists 𝑛0 such that for 𝑛 ≥ 𝑛0 we have 𝑥𝑛 ∈ 𝐻 . Prove that the null sequences N form a normal subgroup of the Cauchy sequences C. Answer: Let C be the group of Cauchy sequences and N the group of null sequences of elements ∈ 𝐺 . Let 𝐹 △ left be the set of all normal subgroups of finite index in G . Clearly 𝐹 △ 𝑙𝑒𝑓𝑡 ⊆ 𝐹 . Every 𝐻 ∈ 𝐹 𝐹 gives rise to a an action of F on the cosets: a homomorphism 𝐹 ⇒ 𝑆𝑦𝑚 ( ) . The group on the right is 𝐻 finite, hence the kernel N of this action is ∈ 𝐹 , and clearly 𝑁 ≤ 𝐻 . For every 𝐻 ∈ 𝐹 , we find 𝑁 ∈ 𝐹 △ left with 𝑁 ≤ 𝐻 . −1 Thus, 𝑥𝑛 is Cauchy only if for every N\in F\triangleleft , there exists 𝑛0 such that 𝑥𝑛 𝑥𝑚 ∈ 𝑁∀𝑛, 𝑚 ≥ 𝑛0 and 𝑥𝑛 is null only if for every 𝑁 ∈ 𝐹 △ left , there exists 𝑛0 such that 𝑥𝑛 ∈ 𝑁∀𝑛, 𝑚 ≥ 𝑛0 𝐺 For 𝑁 ∈ 𝐹 △ left, we have a homomorphism ϕ𝑁: 𝐶 ⇒ : 𝑁 −1 Given 𝑥𝑛 ∈ 𝐶, pick 𝑛0 such that 𝑥𝑛 𝑥𝑚 ∈ 𝑁∀𝑛, 𝑚 ≥ 𝑛0 and set ϕ𝑁(𝑥𝑛 ) = 𝑥𝑛0 𝑁. This is well-defined. By combining all ϕ𝑁 , we obtain a homomorphism ϕ: 𝐶 ⇒ ∏𝑁∈𝐹△𝑙𝑒𝑓𝑡 𝐺/𝑁 We see that ker ϕ = 𝑁 . Therefore, 𝑁 △ left C . Question #18 Let 𝑎1 , 𝑏1 be two real numbers such that 0 < 𝑎1 < 𝑏1 . For 𝑛 ≥ 1 , we define 𝑎𝑛+1 = (𝑎𝑛 𝑏𝑛 )1/2 and 𝑏𝑛+1 = (𝑎𝑛 +𝑏𝑛) 2 a) Prove that the sequence 𝑎𝑛 𝑛 ∈ 𝑁 is monotonically increasing and that the sequence 𝑏𝑛 𝑛 ∈ 𝑁 is monotonically decreasing. b) Show that the sequences 𝑎𝑛 and 𝑏𝑛 are bounded. c) Deduce that the two sequences converge and prove that they converge to the same limit. Answer: Given 𝑎1 , 𝑏1 be two real numbers with 0 < 𝑎1 < 𝑏1 For 𝑛 ≥ 1 , we define 𝑎𝑛+1 = (𝑎𝑛 𝑏𝑛 )1/2 and 𝑏𝑛+1 = 𝑎𝑛 +𝑏𝑛 2 1) Now, 𝑏𝑛+1 = 𝑎𝑛 +𝑏𝑛 2 ≥ (𝑎𝑛 𝑏𝑛 )1/2 = 𝑎𝑛+1 or 𝑏𝑛+1 ≥ 𝑎𝑛+1 for all n . Thus, it implies that 𝑎𝑛+1 = (𝑎𝑛 𝑏𝑛 )1/2 ≥ (𝑎𝑛 𝑎𝑛 )1/2 = 𝑎𝑛 or, 𝑎𝑛+1 ≥ 𝑎𝑛 𝑏𝑛 = 𝑏𝑛 + 𝑏𝑛 𝑎𝑛 + 𝑏𝑛 ≥ = 𝑏𝑛+1 2 2 or, 𝑏𝑛 ≥ 𝑏𝑛+1 Hence, we proved that ∀𝑛 ∈ 𝑁 , 𝑎𝑛 is a monotonically increasing sequence and 𝑏𝑛 is a monotonically decreasing sequence. 2) Using 0 < 𝑎1 < 𝑏1 , we have: 𝑏1 ≥ 𝑏𝑛 ≥ 𝑏𝑛+1 ≥ 𝑎𝑛+1 ≥ 𝑎𝑛 ≥ 𝑎1 > 0 which implies that 𝑎𝑛 is a bounded sequence bounded above by 𝑏1 and 𝑏𝑛 is a bounded sequence bounded below by 𝑏1 . Monotonically increasing sequence is convergent and monotonically decreasing sequence is convergent which implies that both 𝑎𝑛 and 𝑏𝑛 are convergent sequences. Let 𝑎𝑛 converges to a and 𝑏𝑛 converges to b then, 𝑏 = 𝑎+𝑏 2 which gives us a=b Thus, both sequences are convergent to the same limit. Question #19 Please, conclude by the Bounded Convergence Theorem that the sequence is convergent 𝑛𝑒 −𝑛 Answer: Let 𝑎𝑛 = 𝑛𝑒 −𝑛 ⇒ 𝑎𝑛 = 𝑛 𝑒𝑛 𝑎𝑛 min = 𝑛 ∞ =0 For maximum Put 𝑑 𝑑𝑛 [𝑎𝑛 ] = 0 𝑑 𝑛 [ ]=0 𝑑𝑛 𝑒 𝑛 𝑒 𝑛 − 𝑛𝑒 𝑛 =0 (𝑒 𝑛 )2 𝑒 𝑛 (1 − 𝑛1 ) = 0 𝑒 𝑛 = 0, 𝑛 = 1 ∴ 𝑛 = 1, 𝑒 𝑛 > 0, not possible 𝑎𝑛 max = 1 𝑒 Thus, 0 ≤ 𝑎𝑛 ≤ 1 𝑒 Given sequence is bounded 𝑑 𝑛 [𝑎 ] > 0 𝑑𝑛 ⇒ 𝑒 𝑛 (1 − 𝑛) > 0 ⇒ 1−𝑛 > 0 ⇒1>𝑛 𝑛<1 For n<1 sequence is decreasing, thus it is convergent Question #20 Please, find the limit superior and limit inferior of each of the following sequences 𝑛π 1) (𝑥𝑛 ) = (𝑛 cos ( )) , 𝑛 ∈ 𝑁 2 𝑛π 2) (𝑥𝑛 ) = (𝑛 cos 2 ( )) , 𝑛 ∈ 𝑁 2 3) (𝑥𝑛 ) = ((−1)𝑛 − 𝑛 𝑛+1 ),𝑛 ∈ 𝑁 Answer: 𝑛π (1) (𝑥𝑛 ) = (𝑛 cos ( )) , 𝑛 ∈ 𝑁 2 (𝑥4𝑛 ) = (4𝑛 cos ( 4𝑛π )) 2 = 4𝑛 cos 2 π (𝑥4𝑛+1 ) = ((4𝑛 + 1) cos ( (4𝑛 + 1)π )) 2 π = ((4𝑛 + 1) cos (2𝑛π )) 2 = ((4𝑛 + 1) × cos π ) 2 = ((4𝑛 + 1) × 0) = 0 (𝑥4𝑛+2 ) = ((4𝑛 + 2) cos ( (4𝑛 + 2)π )) 2 = ((4𝑛 + 2) cos(2π𝑛 + π)) = ((4𝑛 + 2)(− cos π) = (4𝑛 + 2) (𝑥4𝑛+3 ) = ((4𝑛 + 3) cos ( (4𝑛 + 3)π )) 2 = ((4𝑛 + 3) × cos (2𝑛π + 3π )) 2 = ((4𝑛 + 3) × 0) = 0 Now, lim (𝑥4𝑛 ) = ∞ 𝑛⇒∞ lim (𝑥4𝑛+1 ) = 0 𝑛⇒∞ lim (𝑥4𝑛+2 ) = ∞ 𝑛⇒∞ lim (𝑥4𝑛+3 ) = 0 𝑛⇒∞ Thus, the limit superior of this sequence is ∞ and limit inferior is 0 𝑛π (2) (𝑥𝑛 ) = (𝑛 cos 2 ( )) , 𝑛 ∈ 𝑁 2 (𝑥4𝑛+1 ) = ((4𝑛 + 1) cos 2 ( (4𝑛 + 1)π )) 2 π = ((4𝑛 + 1) cos 2 (2π𝑛 )) 2 π = ((4𝑛 + 1) × cos 2 ( )) 2 = ((4𝑛 + 1) × 0) = 0 Similarly (𝑥4𝑛+2 ) = ((4𝑛 + 2) cos 2 ( (4𝑛+2)π 2 )) = ((4𝑛 + 2) cos 2(2π𝑛 + π)) = ((4𝑛 + 2)(− cos 2 π)) = (−(4𝑛 + 2)) (𝑥4𝑛+3 ) = ((4𝑛 + 3) cos 2 ( (4𝑛 + 3)π )) 2 = ((4𝑛 + 3) × 0) = 0 lim (𝑥4𝑛 ) = ∞ 𝑛⇒∞ lim (𝑥4𝑛+1 ) = 0 𝑛⇒∞ lim (𝑥4𝑛+2 ) = −∞ 𝑛⇒∞ lim (𝑥4𝑛+3 ) = 0 𝑛⇒∞ Thus, limit superior is ∞ and limit inferior is −∞