Mathematics 116 HWK 21 Solutions 8.2 p580

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1 Mathematics 6 HWK Solutios 8. p580 A abbreviatio: iff is a abbreviatio for if ad oly if. Geometric Series: Several of these problems use what we worked out i class cocerig the geometric series, which I ll recap here. The geometric series a + ar + ar + + ar + will coverge if ad oly if r <. If r <, the the sum of the series, which we also wrote as lim S a, will be r. I other words, a + ar + ar + + ar + = a r if r < ad diverges otherwise. Term Test for Divergece: If lim a 0 the a diverges. If lim a = 0, the chace to coverge, but further testig is eeded to decided whether it does or ot. a has a Harmoic Series: = + so the harmoic series diverges. Problem 9, 8., p580. Let a = +. (a) Determie whether the sequece {a } is coverget. (b) Determie whether the series a is coverget. Solutio. (a) Sice lim a = lim + = lim + =, the sequece {a } coverges (to ). (b) From part (a), we kow lim a = 0. Accordig to the Term Test, the series diverges. (We ca also say a = + i this case.) a Page of 7 A. Sotag November, 00

2 Math 6 HWK 8. p580 Sols cotiued Determie whether the series coverges or diverges. If it coverges, fid its sum. Problem, 8., p580. ( ) 5 Solutio. The give series ( ) 5 = is geometric, with iitial term a = 5 = ad commo ratio r =. Sice r = coverges. For the sum, we have <, the series ( ) 5 = 5 = 5 Problem 5, 8., p or or or Solutio. This is a geometric series with a = 8 = 64 ad r = 8. Sice r = 8 = 8 series diverges. It would be appropriate to write 8 + = +. >, this Problem 6, 8., p580. e or (e ) or e + e 4 + e 6 + Solutio. This is geometric with a = e ad r = e. Sice r = <, the series coverges. e Its sum is e = e + e 4 + e 6 + = e e = e Page of 7 A. Sotag November, 00

3 Math 6 HWK 8. p580 Sols cotiued Problem 7, 8., p Solutio. Sice lim + 5 = lim + 5 = 0, the Term Test shows that the give series + 5 diverges. I fact, + 5 = +. Problem 8, 8., p580. Solutio. The give series is a variatio o the harmoic series. Accordig to Theorem 8(i) p579, if the give series did coverge, the the harmoic series = ( ) would also coverge. We kow that the harmoic series diverges, so the give series must also diverge. I fact = +. Problem, 8., p580. [(0.) + (0.) ] Solutio. Use Theorem 8 (ii), which tells us that two coverget series ca be added together term-by-term to form a ew coverget series, ad the sum of the ew series is obtaied by addig the sums of the two origial series. I other words, provided that a ad (a + b ) = a + b are both coverget. I this problem, we re usig a = (0.) = (0.) + (0.) + (0.) + b ad b = (0.) = (0.) + (0.) + (0.) + Each of these series is a coverget geometric series. For the first oe, a = (0.) = 0. ad r = 0.. Page of 7 A. Sotag November, 00

4 Math 6 HWK 8. p580 Sols cotiued For the secod series, a = 0. ad r = 0.. Puttig all this iformatio together ad usig the formula for the sum of a geometric series, we have [(0.) + (0.) ] = (0.) + (0.) = = = Problem, 8., p580. [ ( ) ( )] si si or + [si si( ] ) + [si( ] ) si( ) + [si( ] [ ) si(4 ) + + si( ] ) si( + ) + Solutio. This series telescopes. Writig S N = N [ ( ) ( )] si si, we have + S N = [si si( ] ) + [si( ] ) si( ) + [si( ] [ ) si(4 ) + + si( ] N ) si( N + ) = si si( N + ). Therefore lim S N = lim [si si( )] = si si 0 = si. The give series coverges N N N + ad its sum is si. Problem 5, 8., p Solutio. As i Problem, we ca use Theorem 8(ii). This time a = 6 = ( ) = 6 ( ) = ad b = 6 = ( ) = 6 ( ) = Each of these series is a coverget geometric series. For the first oe, a = ad r =. For the Page 4 of 7 A. Sotag November, 00

5 Math 6 HWK 8. p580 Sols cotiued secod series, a = ad r =. Puttig this iformatio together ad usig the formula for the sum of a geometric series, we have + 6 = = + = + = Problem 6, 8., p Solutio. series diverges: Sice lim = lim 5 + = = 5 0, the Term Test shows that the give Problem 8, 8., p580. l + (See the hit o the HWK assigmet page.) Solutio. The Term Test is of o use here, sice lim examie the partial sums S. We have l S = l + l + l l + + = l + = l = 0. Followig the hit, Therefore lim S = lim l + l + =. =. By defiitio, the, the give series diverges. I fact Page 5 of 7 A. Sotag November, 00

6 Math 6 HWK 8. p580 Sols cotiued Problem, 8., p580. Fid the values of x for which the series values of x, fid the sum. x coverges. For those x Solutio. The give series, or x + x + x + + x + is geometric, with a = x ad r = x. It coverges if ad oly if r <, so it coverges if ad oly if x <. Sice x < x <, the give series coverges if ad oly if < x < [i.e. if ad oly if x belogs to the ope iterval (,)]. For the sum, we have x = x + x + x + + x x + = x = x x provided x <. Problem 4, 8., p580. For those values of x, fid the sum. Fid the values of x for which the series (x + ) coverges. =0 Solutio. The give series, (x + ) = + (x + ) + (x + ) + (x + ) + + (x + ) + =0 is geometric, with a = ad r = (x + ). It coverges iff r <, hece iff (x + ) <. The coditio (x + ) < is equivalet to x + < or x ( ) < or < x <. So the series coverges iff x belogs to the ope iterval (, ), a iterval cetered at. For these values of x the sum is (x + ) = =0 (x + ) = x + Page 6 of 7 A. Sotag November, 00

7 Math 6 HWK 8. p580 Sols cotiued Problem 5, 8., p580. Fid the values of x for which the series sum of the series for those values of x for which it coverges. =0 coverges. Fid the x Solutio. The give series =0 x = + x + x + x + + x + is geometric, with iitial term a = ad commo ratio r = x. It therefore coverges iff r = x <. Sice x < iff x >, the give series coverges iff x >. I other words, it coverges if x <, it diverges if x, ad it coverges if x >. Whe x >, the series sum is =0 x = x = x x Problem 9, 8., p580. S = +. Fid a ad fid Suppose that the th partial sum of a series a. a is give by Solutio. Accordig to the defiitio of series sum, a = lim S = lim + = lim + To fid a formula for a, we ca reaso as follows. For =, we have a = a = S = + = 0. For >, we kow S is the same as S + a, so a = S S = ( ) + ( ) + = + ( ) ( )( + ) = = ( + ) = ( + ) for >. Page 7 of 7 A. Sotag November, 00