# midterm2_prac3solution - Calculus Math 21C Fall 2010...

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Calculus:Math 21C, Fall 2010Solutions to Sample Questions: Midterm 21.Determine the interval of convergence (including the endpoints) for thefollowing power series. State explicitly for what values ofxthe series con-verges absolutely, converges conditionally, and diverges. In each case, specifythe radius of convergenceRand the center of the interval of convergencea.(a)Xn=1(-1)n2nn(x-1)n;(b)Xn=013n+ 1x2n;(c)Xn=01n25n(2x+ 1)n.Solution.We give two different (but related) methods for finding the solution.Either one is fine.Method I.Write the power series asn=0an.(a) We havelimn→∞an+1an= limn→∞(-1)n+12n+1(x-1)n+1/(n+ 1)(-1)n2n(x-1)n/n= limn→∞2(x-1)nn+ 1= 2|x-1|limn→∞nn+ 1= 2|x-1|limn→∞11 + 1/n= 2|x-1|.By the ratio test, the series converges absolutely when2|x-1|<1or|x-1|<1/2.Therefore,a= 1 andR= 1/2.The series converges absolutely if1/2< x <3/2 and diverges if-∞< x <1/2 or 3/2< x <.
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At the endpointx= 3/2 of the interval of convergence, the seriesbecomesXn=1(-1)nn.This is the alternating harmonic series. It converges by the alternatingseries test but does not converge absolutely since the harmonic seriesdiverges. Therefore, the series converges conditionally atx= 3/2.At the endpointx= 1/2 of the interval of convergence, the seriesbecomes-Xn=11n,which a harmonic series. Therefore, the series diverges atx= 1/2.(b) We havelimn→∞an+1an= limn→∞x2(n+1)/(3n+1+ 1)x2n/(3n+ 1)= limn→∞x2(3n+ 1)3n+1+ 1=x2limn→∞3n+ 13n+1+ 1=x2limn→∞1 + 1/3n3 + 1/3n=x23.By the ratio test, the series converges absolutely whenx23<1or|x|<3.Therefore,a= 0 andR=3.The series converges absolutely if|x|<3 and diverges if|x|>3.
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At the endpointsx=±3 of the interval of convergence, the seriesbecomesXn=03n3n+ 1.Sincelimn→∞3n3n+ 1= 1,this series diverges by thenth term test.(c) We havelimn→∞an+1an= limn→∞(2x+ 1)n+1/[(n+ 1)25n+1](2x+ 1)n/[n25n]= limn→∞(2x+ 1)n25(n+ 1)2=|2x+ 1|5limn→∞n2(n+ 1)2=|2x+ 1|5limn→∞1(1 + 1/n)2=|2x+ 1|5.By the ratio test, the series converges absolutely when|2x+ 1|5<1or|x+ 1/2|<5/2.Therefore,a=-1/2 andR= 5/2. The series converges absolutely if-3< x <2 and diverges if-∞< x <-3 or 2< x <.At the endpointsx=-3,x= 2 of the interval of convergence, theseries becomeXn=1(-1)nn2,Xn=11n2.These are absolutely convergentp-series, so the series converges abso-lutely atx=-3,2.