QA #1 Series

Questions and Answers Sheet 1 Series Question #1 Which statement is true for the following series? ∑∞ 𝑘=1 (−1)𝑘 5𝑘+6 1. Cannot be determined 2. Divergent 3. Conditionally convergent 4. Absolutely convergent Answer: Given that ∑∞ 𝑘=1 (−1)𝑘 5𝑘+6 ∞ A series \𝑠𝑢𝑚𝑛=1 𝑎𝑛 is called absolutely Convergent if ∑∞ | | 𝑛=1 𝑎𝑛 is Convergen. ∞ And if ∑∞ 𝑛=1|𝑎𝑛 | is divergent but ∑𝑛=1 𝑎𝑛 is Convergent. ∞ Here ∑𝑛=1 𝑎𝑛 is called Conditionally Convergent. Here 𝑎𝑘 = (−1)𝑘 5𝑘+6 ∞ ∑∞ 𝑘=1|𝑎𝑛 | = ∑𝑘=1 1 5𝑘+6 Limit compoison test. ∞ if ∑∞ 𝑛=1 𝑎𝑛 and ∑𝑛=1 𝑏𝑛 are two series and let lim then ∑∞ 𝑛=1 𝑎𝑛 1 ∴ let 𝑏𝑘 = 𝑘 𝑞𝑘 ∴ lim 𝑞𝑘 ⇒ lim𝑘⇒𝑏𝑘 𝑏𝑘 1 ⇒ ∑∞ 𝑘=1 𝑘 and Since 1 ∑∞ 𝑘=1 𝑘 𝑏𝑛 = 1 non zero finite number both Convergent or both divergent 1 = lim 𝑏𝑘 𝑘⇒𝑏𝑘 and ∑∞ 𝑛=1 𝑏𝑛 𝑎𝑛 5𝑘+6 1 𝑘 = lim 𝑘 1 = ≠ 0 finite number 5 𝑘⇒𝑏𝑘 5𝑘+6 1 ∑∞ 𝑘=1 5𝑘+6 both conv. or div. 1 is divergent therefore ∑∞ 𝑘=1 5𝑘+6 is divergent by limit Compoison test (−1)𝑘 ⇒ ∑∞ 𝑘=1 5𝑘+6 can not be absolutely Convergent. Now. Leibnitz series test. 𝑘 ∑∞ 𝑘=1(−1) 𝑎𝑘 is series, then it is Convergent if it is satisfie following. 1) lim 𝑎𝑘 = 0 𝑘⇒∞ 2) 𝑎𝑘+1 ≤ 𝑎𝑘 , ∀𝑘. ∴ lim 𝑎𝑘 = lim 𝑘⇒∞ 𝑘⇒∞ 1 5𝑘+6 lim 𝑎𝑘 = 0 𝑘⇒∞ 1 1 And 𝑎𝑘+1 − 𝑎𝑘 = 5𝑘+5+6 − 5𝑘+6 5𝑘+6−5𝑘−11 (5𝑘+6)(5𝑘+11) −5 (5𝑘+6)(5𝑘+11) <0 ⇒ 𝑎𝑘+1 < 𝑎𝑘 ∀𝑘 ∈ 𝑁 (−1)𝑘 ⇒ ∑ ∞𝑘 = 1 5𝑘+6 satisfies both property therefore it is Convergent by Leinbnitz test and Conditionally Convergent. So option 3 is correct. Question #2 Select the FIRST correct reason why the given deries converges. A. Convergent geometric series B. Convergent p series C. Comparison (of Limit Comparison) with a geometric or p series D. Converges by alternating series test 1. 𝑃𝑆𝐾 ∑∞ 𝑛=1 2. ∑∞ 𝑛=1 3. ∑∞ 𝑛=1 4. ∑∞ 𝑛=1 sin2 (3𝑛) 𝑛2 (−1)𝑛 ln(𝑒 𝑛) 𝑛 4 cos(𝑛π) cos(𝑛π) ln(4𝑛) 2(4)𝑛 62𝑛 (−1)𝑛 5. ∑∞ 𝑛=1 2𝑛+6 𝑛 𝑛 √ 6. ∑∞ 𝑛=1(−1) 𝑛+4 Answer: 1) 𝑎𝑛 = sin2 (3𝑛) 𝑛2 , 1 𝑛 = 𝑛2 Converges by comparsion with p-series → option (C) 2) 𝑎𝑛 = (−1)𝑛 ln(𝑒 𝑛) 𝑛 4 cos(𝑛π) = (−1)𝑛𝑛 ln(𝑒) 𝑛 4 (−1)𝑛 1 = 𝑛3 → convergent p-series → option (B) Question #3 Determine if the following series is convergent or divergent 1 a) ∑∞ 𝑛=2 𝑛 ln 𝑛 −𝑛 b) ∑∞ 𝑛=0 𝑛𝑒 c) ∑∞ 𝑛=1 d) ∑∞ 𝑛=4 1 √𝑛 1 𝑛7 Answer: 2 Question #4 Find a power series for the function, centered at c, and determine the interval of convergence. 𝑔(𝑥) = 3 2𝑥−1 ,𝑐 = 2 Answer: Geometric Power Series Centered at c: The geometric power series centered at c is a series of the form 𝑎 1−(𝑟−𝑐) 𝑛 = ∑∞ 𝑛=0 𝑎 (𝑟 − 𝑐 ) , |𝑟 − 𝑐 | < 1 where a is the first term and r -c is the common ratio. The function is given by 3 \𝑓(𝑥 ) = 2𝑥−1 , 𝑐 = 2 Writing 𝑓(𝑥 ) in the form 3 = produces 3 2𝑥−1 = 𝑎 1−𝑟 = 2(𝑥−2)+3 3 2 3(1+ (𝑥−2)) 3 1 𝑎 2 3 (1+ (𝑥−2)) = 1−𝑟 2 which implies that 𝑎 = 1 and vr = − 3 (𝑥 − 2) Step 2 So, the power series for f(x) is 3 2𝑥−1 𝑛 = ∑∞ 𝑛=0 𝑎𝑟 𝑛 2 = ∑∞ 𝑛=0 [− 3 (𝑥 − 2)] 2 𝑛 𝑛 = ∑∞ 𝑛=0 (− 3) (𝑥 − 2) 2 4 8 = 1 − 3 (𝑥 − 2) + 9 (𝑥 − 2)2 − 27 (𝑥 − 2)3 + This power series converges when 2 3 𝑣 3 |𝑥 − 2 | < 1 ⇒ |𝑥 − 2 | < 2 3 3 ⇒ −2 < 𝑥 − 2 < 2 3 3 ⇒2−2 < 𝑥 <2+2 1 1 ⇒ 2 < 𝑥 < 32 1 1 which implies that the interval of convergence is ( , 3 ) 2 2 3 Therefore, the power series for the function 𝑓 (𝑥 ) = 2𝑥−1 ,c=2 centered at c is 3 2 2𝑥−1 4 8 = 1 − 3 (𝑥 − 2) + 9 (𝑥 − 2)2 − 27 (𝑥 − 2)3 + Also, the interval of convergence is (1,3) Question #5 Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. 4 + 9 27 3 + 4 + 16 + Answer: Step 1 This is the summation of the series given 4 ∑∞ 𝑛=1 3𝑛−1 = 1 ⋅ 4𝑛−1 Step 2 By simplifying the sequence, we can see that the summation is a geometric series 𝑎𝑟 𝑛−1 where 𝑎 = 4 3 and 𝑟 = 4 since r, the series is convergent Step 3 definition of the limit of the geometric series is 𝑎 1−𝑟 = 4 3 4 1− 𝑎 1−𝑟 = 4 ⋅ 4 = 16 Question #6 ∑∞ 𝑥=1 𝑛+6 𝑛 2 +12𝑛+8 Answer: ∑∞ 𝑥=1 𝑛+6 𝑛 2 +12𝑛+8 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑤ℎ𝑒𝑛 𝑛+6 𝑛(𝑛+12)+8 =0 Question #7 ∑∞ 𝑥=1 1 8𝑛 Answer: ∞ 𝑠𝑢𝑚(𝑥=1) 1/8𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑤ℎ𝑒𝑛 8−𝑛 = 0 Question #8 Which of the following statements are true? ∞ 1.If 𝑎𝑛 and 𝑓 (𝑛) satisfy the requirements of the integral test, then ∑∞ 𝑛=1 𝑎𝑛 = ∫1 𝑓 (𝑥 )𝑑𝑥 . 1 2. The series ∑∞ 𝑛=1 𝑛 𝑝 converges if p > 1 and diverges if 𝑝 ≤ 1 . 3. The integral test does not apply to divergent sequences. Answer: ∞ 1)If 𝑎𝑛 and f(n) satisty the requirements of the integral test, then ∑∞ 𝑛=1 𝑎𝑛 = ∫1 𝑓 (𝑥 )𝑑𝑥 . 1 2)The series ∑∞ 𝑛=1 𝑛 𝑝 converges if p > 1 and diverges if 𝑝 ≤ 1. 3) The integral test can be applied to both covergent and divergent tested Therefore E. Only statements 1 and 2 are true. Question #9 Find the value of x for which the series converges 𝑛 ∑∞ 𝑛=1(𝑥 + 2) Find the sum of the series for those values of x. Answer: 𝑛 Consider the series ∑∞ 𝑛=1(𝑥 + 2) Let 𝑎𝑛 = (𝑥 + 2)𝑛 𝑎𝑛+1 = (𝑥 + 2)𝑛+1 𝑎𝑛+1 = 𝑎𝑛 (𝑥+2)𝑛+1 (𝑥+2)𝑛 =𝑥+2 lim 𝑎𝑛+1 𝑛→∞ 𝑎𝑛 = lim (|𝑥 + 2|) = |𝑥 + 2| 𝑛→∞ Using ratio test, this will converge, when |𝑥 + 2| < 1 −1 < (𝑥 + 2) < 1 −3 < 𝑥 < −1 So, the series is converges at the values lies in the interval, (−3, −1) 𝑛 1 2 3 ∞ And ∑∞ 𝑛=1(𝑥 + 2) = (𝑥 + 2) + (𝑥 + 2) + (𝑥 + 2) + ⋯ + (𝑥 + 2) 𝑥+2 = 1−(𝑥+2) = 𝑥+2 −1−𝑥 Question #10 𝑛 Starting with the geometric series ∑∞ 𝑛=0 𝑥 , find the sum of the series 𝑛−1 | | ∑∞ , 𝑥 <1 𝑛=1 𝑛𝑥 Answer: Consider the geometric series, 𝑛 2 3 ∑∞ 𝑛=0 𝑥 = 1 + 𝑥 + 𝑥 + 𝑥 + ⋯ 1 = 1−𝑥 𝑛−1 | | Find the sum of the series ∑∞ , 𝑥 <1 𝑛=1 𝑛𝑥 𝑛−1 ∑∞ = 1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + ⋯ 𝑛=1 𝑛𝑥 = (1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ ) + (𝑥 + 2𝑥 2 + 3𝑥 2 + 4𝑥 3 + ⋯ ) = = 1 1−𝑥 1 1−𝑥 + 𝑥(1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + ⋯ ) 𝑛−1 + 𝑥 ∑∞ 𝑛=1 𝑛𝑥 1 𝑛−1 𝑛−1 ⇒ ∑∞ − 𝑥 ∑∞ = 1−𝑥 𝑛=1 𝑛𝑥 𝑛=1 𝑛𝑥 1 𝑛−1 (1 − 𝑥 ) ∑∞ = 1−𝑥 𝑛=1 𝑛𝑥 1 𝑛−1 ∑∞ = (1−𝑥)2 𝑛=1 𝑛𝑥 Question #11 Find the sim of each of the following series. 𝑛 1) ∑∞ 𝑛=1 𝑛𝑥 , |𝑥 | < 1 𝑛 2) ∑∞ 𝑛=1 8𝑛 Answer: 𝑛 1) Find the sum of the series ∑∞ 𝑛=1 𝑛𝑥 , |𝑥 | < 1 𝑛 2 3 4 ∑∞ 𝑛=1 𝑛𝑥 = 𝑥 + 2𝑥 + 3𝑥 + 4𝑥 + ⋯ = 𝑥(1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + ⋯ ) 𝑛−1 = 𝑥 ∑∞ 𝑛=1 𝑛𝑥 1 = 𝑥 (1−𝑥)2 𝑥 = (1−𝑥)2 𝑛 2) Find the sum of the series ∑∞ 𝑛=1 8𝑛 ∑∞ 𝑛=1 1 𝑛 𝑛 8𝑛 = ∑∞ 𝑛=1 𝑛 (8) 1 𝑛 Compare with 𝑃𝑆𝐾 ∑∞ 𝑛=1 𝑛𝑥 𝑍𝑆𝐾 , 𝑡ℎ𝑒𝑛𝑃𝑆𝐾𝑥 = 8 𝑥 𝑛−1 ∑∞ = (1−𝑥)2 𝑛=1 𝑛𝑥 1/8 = 1 2 8 (1− ) 1/8 = (7/8)2 1/8 = 49/64 8 = 49 Question #12 Determine whether the geometric series is convergent or divergent. 10 − 4 + 1.6 − 0.64 + ⋯ If it convergent, find the sum. Answer: 𝑎 = 10 −4 𝑟= = 10 −2 5 Now sum: 𝑎 𝑆𝑛 = 1−𝑟 = = 10 2 5 (1+ ) 50 7 Question #13 Let P(k) be a statement that 1 1⋅2 1 1 + 2⋅3 + ⋯ + 𝑘⋅(𝑘+1) = for: The basis step to prove P(k) is that at k = 1, _____ is true. for:Show that P(1) is true by completing the basis step proof. Left side of P(k) and Right side of P(k) for: Identify the inductive hypothesis used to prove P(k). for: Identify the inductive step used to prove P(k+1). Answer: 1 Let the property P(k) be 1 1⋅2 1 1 𝑘 + 2⋅3 + 3⋅4 + ⋯ + 𝑘(𝑘+1) = 𝑘+1 Show that P(k) is true for all integers 𝑘 ≥ 1 using mathematical induction Basis Step: P(k) is true: That is to show that 1 1⋅2 1 = 1+1 The left hand side of the equation is 1 1 1⋅2 1 = 2 and right-hand side is 1 1+1 =2 1 It follows that 2 1 =2 Hence P(1) is true. Show that for all integers 𝑛 ≥ 1 , P(k) is true then P(k+1) is also true: Suppose P(k) is true. Then the inductive hypothesis is 1 1⋅2 1 1 1 + 2⋅3 + 3⋅4 + ⋯ + 𝑘(𝑘+1) Now show that P(k+1) is true. That is to show that 1 1⋅2 1 1 1 1 𝑘+1 + 2⋅3 + 3⋅4 + ⋯ + 𝑘(𝑘+1) + (𝑘+1)(𝑘+2) = (𝑘+1)+1 Or, equivalently that 1 1⋅2 + 1 2⋅3 + 1 1 3⋅4 + ⋯ + (𝑘+1)(𝑘+2) = 𝑘+1 𝑘+2 The left-hand side of P(k+1) is 1 1⋅2 1 1 1 1 + 2⋅3 + 3⋅4 + ⋯ + 𝑘(𝑘+1) + (𝑘+1)(𝑘+2) 𝑘 1 1 1 = 𝑘+1 + (𝑘+1)(𝑘+2) = 𝑘+1 (𝑘 + 𝑘+2) = 1 𝑘+1 ( 1 = 𝑘+1 ( 𝑘 2 +2𝑘+1 𝑘+2 (𝑘+1)2 𝑘+2 ) ) 1 (𝑘+1)2 = 𝑘+1 𝑘+2 𝑘+1 = 𝑘+2 which is right hand side of P(k+1) Hence from the principle of mathematical induction, 1 1⋅2 1 1 1 𝑛 + 2⋅3 + 3⋅4 + ⋯ + 𝑛(𝑛+1) = 𝑛+1 is true, for all integers 𝑛 ≥ 1 Question #14 Test the series for convergence or divergence. ∑∞ 𝑛=0 (−1)𝑛+1 √𝑛+4 Answer: Given: ∑∞ 𝑛=0 (−1)𝑛+1 √𝑛+4 On applying radical rule ∑∞ 𝑛=0 (−1)𝑛+1 √𝑛+4 (−1)𝑛+1 = ∑∞ 𝑛=0 (𝑛+4)1/2 On applying rule exponent rule ∑∞ 𝑛=0 (−1)𝑛+1 √𝑛+4 −1 𝑛+1 ( = ∑∞ 𝑛 + 4) 2 𝑛=0(−1) On simplifying, −1 (𝑛 + 4) 2 = ∑∞ 𝑛=0 (−1)𝑛+1 √𝑛+4 1 √𝑛+4 𝑛+1 = ∑∞ 𝑛=0(−1) 1 √𝑛+4 On applying the alternating series test the function converges (−1)𝑛+1 therefore, ∑∞ 𝑛=0 √𝑛+4 converges Question #15 Using only the definition of Riemann sum and your knowledge of limits, compute the exact area under the curve 2𝑥 2 between 𝑥 = −2 and 𝑥 = 1. Answer: Divide the are in n strips of rectangles with breath: △𝑥 = 1−(−2) 𝑛 3 =𝑛 Now using right hand Riemann sums the area under the curve 2𝑥 2 is: 𝑆𝑛 =△ 𝑥 ∑𝑛𝑖=1 2(−2 + 𝑖 △ 𝑥 )2 3 2 3 = 𝑛 ∑𝑛𝑖=1 2 (−2 + 𝑖 𝑛) 3 = 𝑛 ∑𝑛𝑖=1 2 (4 − = 24 = 24 𝑛 𝑛 ∑𝑛𝑖=1 1 − 12 𝑛 72 72 54 ∑𝑛𝑖=1 𝑖 + 𝑛3 𝑛(𝑛+1) 54 𝑛2 × 𝑛 − 𝑛2 × 9 𝑖 + 𝑛2 𝑖 2 ) 2 1 ∑𝑛𝑖=1 𝑖 2 + 𝑛3 × 𝑛(𝑛+1)(2𝑛+1) 6 1 1 = 24 − 36 (1 + 𝑛) + 9 (1 + 𝑛) (2 + 𝑛) Now to calculate the area decrease the width of the rectangular strips to infidecimal value thereby increasing the number of rectangles to infinity. Finally the required area of the rectangle will be: lim 𝑛→∞ 𝑆𝑛 = lim 𝑛→∞ 1 24 − 36 (1 + 𝑛) 1 = 24 − 36(1 + 0) + 9(1 + 0)(2 + 0) = 24 − 36 + 18 =6 Thus, area under the curve 2𝑥 2 is 6 Question #16 Find the coefficient of 𝑥 5 𝑦 8 𝑖𝑛 (𝑥 + 𝑦)13 1 𝑛 (1+ ) (2 + 𝑛) Answer: Binomial theorem n n ( x + y)n =  ( ) x n− j y j j =0 j We are interested in the term 𝑥 5 𝑦 8 in (𝑥 + 𝑦)13 𝑛 = 13 𝑗=8 The coefficient of this term is then n 13 ( )=( ) j 8 n 13 13! 13! ( )=( )= = = 1287 j 8 8!(13 − 8)! 8!5! Result: 1287 Question #17 Write out he first four terms of the Maclaurin series of f(x) if 𝑓 (0) = 2, 𝑓(0) = 3, 𝑓 (0) = 4, 𝑓 (0) = 12 Answer: Maclaurin series: 𝑓(𝑥 ) = ∑∞ 𝑛=0 𝑓(𝑛)(0) 𝑛! 𝑥 𝑛 = 𝑓 (0) + 𝑓 ′ (0)𝑥 + 𝑓′′ (0) 2! 𝑥2 + 𝑓′′′ (0) 3! 𝑥3 + 𝑓 (4) (0) 4! 𝑥4 + ⋯ Here given f(0)=2,f'(0)=3,f''(0)=4 and we need to find out the first 4 terms of the Maclaurin Series 𝑓(𝑥 ) = ∑∞ 𝑛=0 𝑓(𝑛) (0) 𝑛! 𝑥 𝑛 = 𝑓 (0) + 𝑓 ′ (0)𝑥 + 𝑓′′ (0) 2! 𝑥2 + 𝑓′′′ (0) 3! 𝑥3 + ⋯ We need to find 4 terms 4 𝑓(𝑥 ) = 2 + 3𝑥 + 2 𝑥 2 + 12 6 𝑥3 𝑓(𝑥 ) = 2 + 3𝑥 + 2𝑥 2 + 2𝑥 3 Result: 𝑓(𝑥 ) = 2 + 3𝑥 + 2𝑥 2 + 2𝑥 3 Question #18 Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. 10 − 2 + 0.4 − 0.008 + ⋯ Answer: Geometric series: 𝑎 𝑛 ∑∞ 𝑛=0 𝑎 ⋅ 𝑟 = 1−𝑟 Where r and a are constants If |𝑟| < 1, then the series converges to 𝑎 1−𝑟 10 − 2 + 0.4 − 0.08 … 1 1 2 1 3 10 + 10 ⋅ (− 5) + 10 ⋅ (− 5) + 10 ⋅ (− 5) … 1 𝑛 ∑∞ 𝑛=0 10 ⋅ (− 5) 1 This is a geometric series with common ration 𝑟 = − 5 and Initial Term 𝑎 = 10 1 Since |𝑟| = 5 < 1, the given geometric series converges. Sum of the geometric series is 𝑎 10 𝑆 = 1−𝑟 = 1 5 1−(− ) = 10 1 5 1+ = The series converges to 10 6 5 = 50 6 = 25 3 25 3 Question #19 Find a power series representation for the function and determine the interval of convergence. 𝑓(𝑥 ) = 2/3 − 𝑥 Answer: 2 𝑓(𝑥 ) = 3−𝑥 Divide Numerator and denominator by 3. To get 2 3 𝑓 (𝑥 ) = 𝑥 3 1− Note that the sum of the geometric series with Initial term a and common ratio r is 𝑛 𝑆 = ∑∞ 𝑛=0 𝑎𝑟 = 𝑎 1−𝑟 The given function can be interpred as 2 3 1− 𝑎 = 1−𝑟 𝑥 3 2 Therefore, we can say that f(x) is a sum of a geometric series with initial term 𝑎 = 3 and common ratio 𝑥 𝑟=3 Therefore, 2 𝑥 𝑛 3 3 ∞ ∞ 𝑛 𝑓(𝑥 ) = ∑∞ 𝑛=0 𝑎𝑟 = ∑𝑛=0 ( ) ( ) = ∑𝑛=0 ( 2 3𝑛+1 ) ⋅ 𝑥𝑛 This is the power series representation of f(x) We know that the geometric series converges when |𝑟| = |⍁{𝑥}{3}| < 1 |𝑥 | < 3 Intterval of converges is (−3,3) Radius of converges is 3 Question #20 Find a power series for the function, centered at c, and determine the interval of convergence. 𝑔(𝑥 ) = 3/2𝑥 − 1, 𝑐 = 2 Answer: Geometric Power Series Centered at c: The geometric power series centered at c is a series of the form 𝑎 1−(𝑟−𝑐) 𝑛 = ∑∞ 𝑛=0 𝑎 (𝑟 − 𝑐 ) , |𝑟 − 𝑐 | < 1 where a is the first term and r-c is the common ratio. The function is given by 3 𝑓(𝑥 ) = 2𝑥−1 , 𝑐 = 2 𝑎 Writing f(x) in the form 3 2𝑥−1 = = 1−𝑟 produces 3 = 2(𝑥−−2)+3 3 2 3 3(1+ (𝑥−2)) 1 𝑎 2 3 (1+ (𝑥−2)) = 1−𝑟 2 which implies that a=1 and 𝑟 = − 3 (𝑥 − 2) . So, the power series for f(x) is 3 2𝑥−1 𝑛 = ∑∞ 𝑛=0 𝑎𝑟 𝑛 2 = ∑∞ 𝑛=0 [− 3 (𝑥 − 2)] 2 𝑛 𝑛 = ∑∞ 𝑛=0 (− 3) (𝑥 − 2) 2 4 8 = 1 − 3 (𝑥 − 2) + 9 (𝑥 − 2)2 − 27 (𝑥 − 2)3 + ⋯ This power series converges when 2 3 |𝑥 − 2 | < 1 ⇒ |𝑥 − 2 | < 3 3 2 3 ⇒ −2 < 𝑥 − 2 < 2 1 1 ⇒ 2 < 𝑥 < 32 1 1 which implies that the interval of convergence is (2 , 3 2) The power series converges when 2 3 |𝑥 − 2 | < 1 ⇒ |𝑥 − 2 | < 3 3 2 3 ⇒ −2 < 𝑥 − 2 < 2 3 3 2−2< 𝑥 <2+2 1 1 2 2 ⇒ <𝑥<3 1 1 which implies that the interval of converges is (2 , 3 2) 3 Therefore, the power series for the function 𝑓 (𝑥 ) = 2𝑥−1 , 𝑐 = 2 centered at c is 3 2𝑥−1 2 4 8 = 1 − 3 (𝑥 − 2) + 9 (𝑥 − 2)2 − 27 (𝑥 − 2)3 + ⋯ Also, the interval of convergence is (1,3). Final answer: 3 2𝑥−1 2 4 8 3 9 27 = 1 − (𝑥 − 2) + (𝑥 − 2 )2 − (𝑥 − 2)3 + ⋯