Name: Diana ClarkUniversity ID: 0644492Thomas Edison State UniversityCalculus II (MAT-232)Section no.: WA5Semester and year: September 2020Written Assignment 5Answer all assigned exercises, and show all work. Each exercise is worth 5 points.*Submitting a graph is not required; however, you are encouraged to create one for your own benefit and to include (or describe) one if possible.Section 8.64.Determine the radius and interval of convergence.02kkkkxlimk →∞|ak+1ak|=limk →∞|(k+1)xk+1(2k)2k+1(xk)|= limk →∞|x2(k+1k)|= |x|2<1x=2, ∑k=1∞k2k2k=∑k=1∞k=∑k=1∞aklimk →∞ak≠0, diverges. x=−2, diverges Interval convergence (−2,2), radius of convergence is r=210.Determine the radius and interval of convergence.241(32)kkxkWA 5, p. 1
limk →∞|(3x+2)k+1(k+1)2k2(3x+2)k|¿limk →∞|k2(3x+2)(k2+2k+1)|¿limk →∞(3x+2)−1<3x+2<1¿−1<x←13Interval of convergence is (−1,−13)and radius is R=1312.Determine the radius and interval of convergence.1( 1)(31)kkkxkuk=1√k(3x−1)k,uk+1=1√k+1(3x−1)k+1limk →∞|uk+1uk|=limk→ ∞|1√k+1(3x−1)k+11√k(3x−1)k|¿limk →∞|(3x−1)√k√k+1|¿|3x−1|limk →∞√11+1k¿|3x−1|(1)−1<(3x−1)<1= 0<x<23Interval of convergence is (0,23), radius is R=2316.Determine the radius and interval of convergence.2212( !)(2 )!kkkxkWA 5, p. 2
limk →∞|[(k+1)!]2(2k+2)!x2k+2(x(2k+1)!(k!)2x2k+1)|= limk →∞|(k+1)22k+2(1)||x|= limk →∞|(k+1)2(2k+2)||x|limk→ ∞12|1(k+1)||x|=0Converges for every xRadius of convergence will be R=∞, interval of convergence is (−∞ ,∞)24.Determine the interval of convergence and the function to which the given power series converges.034kkx∑k=0∞3(x4)k=3(1+x4+(x4)2+…)For each fixed x, this is a geometric series with common ratio r=x4This converges whenever |r|=|x4|<1and diverges |x|>4.