Practice Problems for Midterm 2 1 Compute the radius of convergence of the series X 4n n 1 n x 3 n Solution We will determine the radius of convergence by applying the ratio test We have 4n 1 n 1 n 1 x 3 lim 4n n n n x 3 lim 4 x n 1 n 1 3 1 n lim 4 x 3 n n 1 n 4 x 3 n since limn n 1 1 By the ratio test the series diverges if 4 x 3 1 and converges when 1 4 x 3 1 x 3 4 so the radius of convergence is 14 2 Compute the radius of convergence of the series X xn 2n n2 n 1 Solution We have xn 1 lim n 2n 1 n 1 2 xn 2n n2 1 x n 1 2 lim 1 n 2 n2 2 n x lim n 2 n 1 x 2 1 P 4n n 1 n x 3 n 2 n 2 1 By the ratio test the series diverges if since limn n 1 x 2 1 and converges when x 1 x 2 2 so the radius of convergence is 2 3 Compute the radius of convergence of the series X 2xn n 1 3n n2 Solution We have 2xn 1 lim n 3n 1 n 1 2 2xn 3n n2 1 x n 1 2 lim 1 n 3 n2 2 x n lim n 3 n 1 x 3 n 2 since limn n 1 1 By the ratio test the series diverges if x 3 1 and converges when x 1 x 3 3 so the radius of convergence is 3 4 Calculate the sum X 1 n 2n 1 n 1 3n Solution This is a constant multiple of a geometric series Factoring out the 2 we obtain X X 1 n 2n 1 n 2n 1 2 n n 3 3 n 1 n 1 3 Note 1 n 2n 3n n 2 3 so this is equal to 2 n X 2 n 1 We factor out 2 3 1 2 3 3 and then apply the geometric series formula since n 2 X 2 4 1 2 3 n 0 3 3 1 32 4 3 1 23 4 5 5 Calculate the sum X 1 n 3n 1 n 3 4n Solution Factoring out the 3 we obtain X 1 n 3n 1 n 3 Note 1 n 3n 4n 4n 3 X 1 n 3n 4n n 3 n 3 4 so this is equal to 3 n X 3 n 3 4 4 3 27 We factor out 3 4 64 and then apply the geometric series formula since 3 4 1 n 27 X 3 1 81 3 64 n 0 4 64 1 34 81 64 1 34 81 112 6 Calculate the sum X 2n 1 n n 0 3n Solution First we distribute the sum to obtain X 2n 1 n X 2n X 1 n n n 3 3 3n n 0 n 0 n 0 n n X X 2 1 3 3 n 0 n 0 Since 23 1 and 1 3 1 the geometric series formula tells us that the above is equal to 1 1 1 1 2 1 1 4 1 3 1 3 3 3 3 3 4 15 4 7 Calculate the sum X 5n 5 n 1 n 3 n 4 n 3 5 Solution This is a telescoping series For M 3 let M M X X 5 n 1 n 1 5n n SM 5 n 3 n 4 n 3 n 4 n 3 n 3 denote the M th partial sum The first four partial sums are 3 4 S3 5 6 7 3 4 4 5 S4 5 6 7 7 8 3 4 4 5 5 6 S5 5 6 7 7 8 8 9 3 4 4 5 5 6 6 7 S6 5 6 7 7 8 8 9 9 10 Notice all terms cancel except the first positive term 63 that comes M 1 from plugging in n 3 and the last negative term M 4 that comes from plugging in n M Therefore 3 M 1 SM 5 6 M 4 1 Since limM M M 4 1 we have X 5n 5 n 1 3 5 lim SM 5 1 M n 3 n 4 6 2 n 3 8 Calculate the sum X ln n 1 n 1 n 2 ln n 2 n 3 Solution This is a telescoping series For M 3 let M X ln n 1 ln n 2 SM n 2 n 3 n 1 6 denote the M th partial sum The first four partial sums are ln 2 3 ln 2 S2 3 ln 2 S3 3 ln 2 S4 3 S1 ln 3 4 ln 3 ln 3 ln 4 4 4 5 ln 3 ln 3 ln 4 ln 4 ln 5 4 4 5 5 6 ln 3 ln 3 ln 4 ln 4 ln 5 ln 5 ln 6 4 4 5 5 6 6 7 Notice all terms cancel except the first positive term ln32 that comes 2 that from plugging in n 3 and the last negative term ln M M 3 comes from plugging in n M Therefore SM ln 2 ln M 2 3 M 3 2 Using L hopital s rule we find that limM ln M M 3 0 so we have X ln n 1 ln n 2 ln 2 lim SM M n 2 n 3 3 n 1 9 Consider the following sequences Circle the correct answers below and show all supporting work n ln n o n 1 o n o 2 n n 1 I 1 II III ln n 1 n 1 2n2 1 n 1 n n Sequence I converges diverges Solution Recall the following theorem Suppose an n 1 is a se quence of nonnegative numbers If an n 1 converges to 0 then the n sequence 1 n an n 1 also converges to 0 Otherwise 1 an n 1 diverges We have n2 1 0 2n2 1 7 for n 1 Since n2 1 1 lim n 2n2 1 2 is nonzero it follows from the theorem that the sequence diverges Sequence II converges diverges is an indeterminate form of type Solution The limit limx ln x x By L hopital s rule we have 1 ln x 1 x lim lim lim 2 0 x x x x x x Therefore ln n 0 n n so the sequence converges lim Sequence III converges diverges Solution We have limn n1 0 Since limx 0 ln x this shows that 1 lim ln n n so the sequence diverges 10 Consider the following sequences Circle the correct answers below and show all supporting work n o n o n o 2 2 nn 1 n nn 1 I 1 II 1 III 1 n ln n n 1 n 1 2n 2n2 n 1 Sequence I converges …