QA #3 Series

Questions and Answers Sheet 3 Series Question #1 Tell whether the series converge. If it converges, find the sum. ∑(𝑛 = 0)𝑜 𝑜(𝑝𝑖/2)𝑛 Answer: ∑(𝑛 = 0)𝑜 𝑜(𝑝𝑖/2)𝑛 𝑎 = 1 = (𝑝𝑖/2)2 /(𝑝𝑖/2) = 𝑝𝑖/2 = 1.57 > Thus, the series diverges. Question #2 Tell whether the series converge. If it converges, find the sum. ∑(𝑛 = 1)𝑜 𝑜(3/7)𝑛 Answer: ∑(𝑛 = 1)𝑜 𝑜(3/7)𝑛 𝑎 = 3/7 < 1 , so the series converges. Sum: (3/7)/(1– 3/7) = = 3/7 𝑥𝑥 7/4 = 3/4 Question #3 Tell whether the series converges or diverges, and find sum, if it converges. ∑(𝑛 = 3)𝑜 𝑜(1/(𝑛– 2)– 1/𝑛) Answer: ∑(𝑛 = 3)𝑜 𝑜(1/(𝑛– 2)– 1/𝑛) ∑(𝑛 = 3)𝑜 𝑜1/(𝑛𝑥𝑥(𝑛 − 2)) This is not geometric series 𝑆𝑛 = 1 + 1/2 = 3/2𝑎𝑠𝑛 → 𝑜𝑜. Thus, series converges. Question #4 Please justify whether the series converges or diverges, and find sum, if it converges. ∑(𝑛 = 1) 𝑜𝑜 3/𝑛2. Answer: ∑(𝑛 = 1) 𝑜𝑜 3/𝑛2 ∑(𝑛 = 1) 𝑜𝑜 1/𝑛𝑝 – if 𝑝 > 1𝑍 – series converges, if 𝑝 < 1 , series diverges. Therefore, series converges by p-test. Question #5 Tell whether the series converges or diverges. ∑(𝑘 = 1)𝑜 𝑜5/(𝑘 + 4)4 Answer: ∑(𝑘 = 1)𝑜 𝑜5/(𝑘 + 4)4 Put 𝑘 + 4 ⇒ 𝑛 So, ∑(𝑛 = 5)𝑜 𝑜5/𝑛4 From p-series test: For series ∑(𝑛 = 1)𝑜 𝑜1/𝑛𝑝 , if 𝑝 > 1, series converges. So, for series in equation 1 𝑃 = 4, so this series converges The series is p-series with 𝑝 = 4 and it converges. Question #6 Justify if the series is convergent: ∑(𝑛 = 1)𝑜 𝑜𝑒 (−𝑛) Answer: ∑(𝑛 = 1)𝑜 𝑜𝑒 (−𝑛) = = ∑(𝑛 = 1)𝑜 𝑜1/𝑒 𝑛 Here, common ratio, 𝑎 = 1/𝑒 So, 𝑆 = (1/𝑒)/(1– 1/𝑒) = (1/𝑒)/((𝑒– 1)/𝑒) 𝑆 = 1/(𝑒– 1). This series converges. Question #7 For which z’s does the infinite series converge? To what function does it converge? 1 + 𝑧/2 + 𝑧 2 /4 + 𝑧 3 /8 + ⋯ Answer: 1 + 𝑧/2 + 𝑧 2 /4 + 𝑧 3 /8 + 𝑍𝑆𝐾 … (geometric series) So, = 1/(1 − (𝑧/2)) = 1/((2– 𝑧)/2) So, = 2/(2– 𝑧) This series converges if |𝑧| < 2 and it converges 2/(2 − 𝑧) . Question #8 Simplify (𝑠𝑒𝑐θ − 𝑡𝑎𝑛θ)(1 + 𝑠𝑖𝑛θ) Answer: (𝑠𝑒𝑐θ − 𝑡𝑎𝑛θ)(1 + 𝑠𝑖𝑛θ) = = 𝑠𝑒𝑐 (θ)𝑠𝑖𝑛(θ) − 𝑡𝑎𝑛(θ) − 𝑡𝑎𝑛(θ)𝑠𝑖𝑛(θ) = 1 𝑠𝑖𝑛θ 𝑠𝑖𝑛θ = 𝑐𝑜𝑠θ + 𝑐𝑜𝑠θ − 𝑡𝑎𝑛θ − 𝑐𝑜𝑠θ 𝑠𝑖𝑛θ = [:′ 𝑠𝑒𝑐θ = = 1 𝑐𝑜𝑠θ 1 = 𝑠𝑖𝑛 2 θ 1−𝑠𝑖𝑛 2 θ 𝑐𝑜𝑠θ 𝑐𝑜𝑠 2 θ 𝑐𝑜𝑠θ 𝑠𝑖𝑛θ 𝑎𝑛𝑑𝑡𝑎𝑛θ = 𝑐𝑜𝑠θ] + 𝑡𝑎𝑛θ − 𝑡𝑎𝑛θ − = 𝑐𝑜𝑠θ − = 1 𝑐𝑜𝑠θ 𝑐𝑜𝑠θ 𝑠𝑖𝑛 2 θ 𝑐𝑜𝑠θ = = = = 𝑐𝑜𝑠 [:′ 𝑠𝑖𝑛2 θ + 𝑐𝑜𝑠 2 θ = 1] Answer: (𝑠𝑒𝑐θ − 𝑡𝑎𝑛θ)(1 + 𝑠𝑖𝑛θ) = 𝑐𝑜𝑠 Question #9 Compute each of these double sums: ∑2𝑖=1 ∑3𝑖+1(𝑖 + 𝑗) Answer: Compute this double sum by first computing the inner sum followed by the outer sum. ∑2𝑖=1 ∑3𝑗=1(𝑖 + 𝑗) = ∑2𝑖=1(𝑖 + 1 + 𝑖 + 2 + 𝑖 + 3) = 3 ∑2𝑖=1(𝑖 + 2) = 3(1 + 2 + 2 + 2) = 21 Question #10 Compute each of these double sums: = ∑2𝑖=1(𝑖 + 1 + 𝑖 + 2 + 𝑗 + 3) Answer: Compute this double sum by first computing the inner sum followed by the outer sum. = ∑2𝑖=0 ∑3𝑗=0(2𝑖 + 3𝑗) = ∑2𝑖=0(2𝑖 + 0 + 2𝑖 + 3 + 2𝑖 + 6 + 2𝑖 + 9) = ∑2𝑖=0(8𝑖 + 18) = 8(0) + 18 + 8(1) + 18 + 8(2) + 18 = 78 Question #11 Write an equivalent series with the index of summation beginning at = 1. 𝑛 𝑛 ∑∞ 𝑛=0(−1) + 1(𝑛 + 1)𝑥 Answer: This equation is given in the series. Add 1 to both sides and introduce k. Subtract 1 from both sides. 𝑛 = 0 , so 𝑛 + 1 = 1 = 𝑘 . Therefore, 𝑛 = 𝑘 − 1 In the original series, plug in 𝑘 − 1 for n [(𝑘−1)+1] [( ∑∞ 𝑘 − 1) + 1] ⋅ 𝑥 𝑘−1 𝑘=1(−1) Simplify. Since k is a dummy variable, you can replace it with any other variable. Let n=k 𝑘 𝑘−1 𝑛 𝑛−1 ∑∞ = ∑∞ 𝑘=1(−1) 𝑘 ⋅ 𝑥 𝑛=1(−1) 𝑛 ⋅ 𝑥 𝑛 𝑛−1 Result: ∑∞ 𝑛=1(−1) 𝑛 ⋅ 𝑥 Question #12 Write an equivalent series with the index of summation beginning at = 1. 𝑛 𝑛 ∑∞ 𝑛=0(−1) + 1(𝑛 + 1)𝑥 Answer: The Taylor series of ln 𝑥 around c=1. We know from the definition of Taylor series of the function f 𝑓(𝑥 ) = 𝑓 (𝑥)|𝑥=𝑐 + 𝑓′ (𝑥)|𝑥=𝑐 1! (𝑥 − 𝑐 ) + 𝑓′′ (𝑥)|𝑥=𝑐 2! (𝑥 − 𝑐 )2 + 𝑓′′′ (𝑥)|𝑥=𝑐 3! (𝑥 − 𝑐 )3 + ⋯ Now the values of differential functions at given point 1 𝑓(1) = ln(1) = 0, 𝑓 ′ (1) = |𝑥=1 = 1 𝑥 1 𝑓 ′′ (1) = − 𝑥 2 2 𝑥=1 𝑛( 6 = −1, 𝑓 ′′′ (1) = 𝑥 3 |𝑥=1 = 2𝑓 ′′′′ (1) = − 𝑥 4 |𝑥=1 = −6 𝑓 1) = (−1)𝑛+1 (𝑛−1)! 𝑥𝑛 |𝑥=1 = (−1)𝑛 ((𝑛 − 1)!) Thus the Taylor series expression become 1 1 1 1 𝑓(𝑥 ) = 0 + (𝑥 − 1) − 2! (𝑥 − 1)2 + 2! 3! (𝑥 − 1)3 − 3! 4! + (−1)𝑛+1 (𝑛 − 1)! 𝑛! (𝑥 − 1)𝑛 Or we can write as 1 1 1 1 f(𝑥 ) = (1 − 𝑥 ) − 2 (𝑥 − 1)2 + 3 (𝑥 − 1)3 − 4 (𝑥 − 1)4 + (−1)𝑛+1 (𝑛 − 1)! 𝑛! (𝑥 − 1)𝑛 − ⋯ Question #13 Use the summation formulas to rewrite the expression withous summatiom notation. ∑𝑛𝑖=1 4𝑖+5 𝑛2 Use the result to find the sums for n=10, 100. Answer: ∑𝑛𝑖=1 𝑖 = 𝑛(𝑛+1) 2 ∑𝑛𝑖=1 𝑘 = 𝑛. 𝑘, where k=constant for 𝑛 = 10 ∑10 𝑖=1 = 4𝑖+5 (10)2 1 10 = 100 [4 ∑10 𝑖=1 𝑖 + ∑𝑖=1 5] 1 10(10 + 1) []4 + 10  5] = 2.7 100 2 for n=100 ∑100 𝑖=1 4𝑖+5 1 (100)2 100 ] = (100)2 [4 ⋅ ∑100 5 𝑖=1 𝑖 + ∑ 𝑖 = 1 1 100(100+1) = (100)2 [4 2 + (5 × 100)] = 2.07 Question #14 Determine the sum of the following series. ∑∞ 𝑛=1 ( 1𝑛+9𝑛 11𝑛 ) Answer: We have to find sum of series. Series is given as: ⇒ ∑∞ 𝑛=1 ( 1𝑛+9𝑛 11𝑛 ) We will use geometric series test to find sum of series . Geometric series test is given below: 𝑛−1 ⇒ ∑∞ 𝑛=1 𝑎𝑟 ⇒ 𝑖𝑓|𝑟| < 1 (series converges) ⇒∑ 𝑎 = 1−𝑟 (infinite gp) With the help of geometric series we will find sum . Work is shown below: ⇒ ∑∞ 𝑛=1 ( 1𝑛+9𝑛 11𝑛 ) 1 𝑛 9 ⇒ ∑∞ 𝑛=1 (11) + (11) 1 𝑛 ⇒ ∑∞ 𝑛=1 (11) = 1 𝑛 𝑛 1 11 1 1− 11 1 ⇒ ∑∞ 𝑛=1 (11) = 10 9 𝑛 ⇒ ∑∞ 𝑛=1 (11) = 9 𝑛 9 11 9 1− 11 9 ⇒ ∑∞ 𝑛=1 (11) = 2 1 𝑛 9 𝑛 ⇒ ∑∞ 𝑛=1 ( ) + ( ) = 11 11 1 𝑛 9 𝑛 ⇒ ∑∞ 𝑛=1 (11) + (11) = 1 10 + 9 2 23 5 Question #15 Which equation could be used to calculate the sum of the geometric series? 1 3 2 4 8 16 + 9 + 27 + 21 + 243 Answer: The formula for the nth partial sum of a geometric sequence is, 𝑎1 (1−𝑟 𝑛) 𝑆𝑛 = ,𝑟 ≠ 1 1−𝑟 where, 𝑆𝑛 is the sum of GP with n terms 𝑎1 is the first term r is the common ratio n is the number of terms The sum of the geometric series given is, 1 3 2 4 8 16 + 9 + 27 + 21 + 243 There are 5 terms in the series and the common ratio of the series is found by dividing any term by the previous term. Suppose, divide the second term by the first term and the common ratio is obtained as, 2 1 𝑟=9÷3 2 =3 Therefore, r is not equal to 1 and the sum of the given series is, 𝑆5 = 1 2 5 (1−( ) ) 3 3 2 (1− ) 3 Question #16 Find whether the series diverges and its sum: 𝑛+1 ∑∞ 𝑛=1(−1) 3 5𝑛 n n n n n n (1 − x)n =  ( )1n−k (− x)k ][( )1n (− x)0 + ( )1n−1 (− x) + ( )1n−2 (− x) 2 + ( )1n−3 (− x)3... 0 1 2 3 k =0 k Answer: Notice that (−1)𝑛+1 3 5𝑛 = −3 (−1)𝑛 5𝑛 −1 𝑛 = −3 ( 5 ) 𝑎𝑟 𝑘 Since ∑∞ 𝑘=1 𝑎𝑟 = 1−𝑟 (if |𝑟 | < 1 ) ∑∞ 𝑛=1 −1 𝑛 −3( 5 ) = −1 5 −1 1− 5 −3⋅ = 3 5 6 5 1 =2 and the sum converges because | −1 1 |=  1 5 5 Question #17 Use linear approximation (or differentials) to estimate: √99.2 What am I supposed to do with this? I am not given x or dx Answer: Ue Taylor series for √𝑥 about 𝑥 = 100 . The reason to expand the Taylor series about 100 is that 100 is the closest square to 99.2. 𝑓(𝑥 ) = 𝑓 (100) + 𝑓 ′ (100)(𝑥 − 100) + higher order terms Hence, 1 (99.2−100) √99.2 ≈ √100 + 2 √100 = 10 − 0.04 = 9.96 Question #18 Determine whether the geometric series is convergent or divergent. 10 − 4 + 1.6 − 0.64 + ⋯. If it's convergent find its sum. Answer: Step 1 Let 𝑆 = 10 − 4 + 1.6 0.64 + ⋯ Given that S is a geometric series, The known fact is that a geometric series 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ is convergent 𝑖𝑓 𝑓 |𝑟| < 1. By comparing the given series with its general form, 𝑎 = 10 , 𝑟 = −4 10 = −0.4 . |𝑟| = |−0.4| < 1 . This implies that the series 10 − 4 + 1.6 − 0.64 + ⋯ is convergent. Step 2 The known fact is that a geometric series 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ - is equal to 𝑎 1−𝑟 (𝑟 < 1) . 10 − 4 + 1.6 − 0.64 ± ⋯ = (10 + 1.6 + 0.256 + ⋯ ) − (4 + 0.64 + ⋯ ) ? a = 10, r = 0.16  10 4 =( )−( ) 1 − 0.16 1 − 0.16  a = 4, r = 0,16  10−4 = (1−0.16) 6 = (0.84) = 7.1428571428571429 Question #19 Binomial expansion of (1 − 𝑥 )𝑛 We've been given with following binomial expansion (𝑥 + 1)𝑛 = 1 + 𝑛𝑥 + 𝑛(𝑛−1) 2! 𝑥2 + 𝑛(𝑛−1)(𝑛−2) 3! 𝑥3 … How can we get the formula of (1 − 𝑥 )𝑛 Answer: Well, as I understand it, we could write the binomial expansion as: n n n n n n (1 − x)n =  ( )1n−k (− x)k ][( )1n (− x)0 + ( )1n−1 (− x) + ( )1n−2 (− x) 2 + ( )1n−3 (− x)3... 0 1 2 3 k =0 k which simplifies to 1 − 𝑛𝑥 + 𝑛(𝑛−1) 2! ⋅ 𝑥2 − 𝑛(𝑛−1)(𝑛−2) 3! ⋅ 𝑥3 … Which is the answer everyone else has given. Question #20 How to find the Maclaurin series for 𝑒 𝑥 ? Answer: The Maclaurin series is obtained by the Power Series: 𝑥2 𝑥 𝑥3 𝑓(𝑥 ) = 𝑓 (0) + 𝑓 ′ (0) 0! + 𝑓 ′ (0) 2! + 𝑓 ′ (0) 3! + ⋯ As we have 𝑓(𝑥 ) = 𝑒 𝑥 , then: 𝑓 (𝑥 ) = 𝑒 𝑥 ⇒ 𝑓 (0) = 1 𝑓 ′ (𝑥 ) = 𝑒 𝑥 ⇒ 𝑓 ′ (0 ) = 1 𝑓 ′′ (𝑥 ) = 𝑒 𝑥 ⇒ 𝑓 ′′ (0) = 1 𝑓 3 (𝑥 ) = 𝑒 𝑥 ⇒ 𝑓 3 (0 ) = 1 𝑓 𝑛 (𝑥 ) = 𝑒 𝑥 ⇒ 𝑓 𝑛 (0 ) = 1 So, the Maclaurin series is: 𝑥 𝑥2 𝑥3 𝑥4 𝑥𝑛 𝑒 𝑥 = 1 + 1 0! + 1 2! + 1 3! + 1 4! + ⋯ + 1 𝑛!