Department of Mathematical and Statistical Sciences University of Alberta

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1 MATH 4 (R) Wier 008 Iermediae Calculus I Soluios o Problem Se # Due: Friday Jauary 8, 008 Deparme of Mahemaical ad Saisical Scieces Uiversiy of Albera Quesio. [Sec.., #] Fid a formula for he geeral erm a of he sequece assumig ha he paer of he firs few erms coiues. { 4, 9, 6, 45,... } Soluio: We oe ha his is a aleraig sequece where he umeraor icreases by sarig wih = ad he deomiaor is he square of + sarig wih =. Hece he geeral formula is a = ( ), =,,.... ( + ) Quesio. [Sec.., #] Deermie wheher he sequece a = ( ) If i coverges, fid he i. + + coverges or diverges. Soluio: Firs oe ha a = + + = + + = =, ad he sequece a, a 4,... approaches whereas he sequece a, a,... approaches (because of he ( ) ). I oher words, he sequece a oscillaes bewee he values ad as. Hece, a does o exis, ad he sequece {a } diverges. Quesio. [Sec.., #6] Deermie wheher he sequece a = arca coverges or diverges. If i coverges, fid he i. Soluio: As,, ad x a x = x a x = π implies so ha {arca } coverges. a = π,

2 Quesio 4. [Sec.., #4] Deermie wheher he sequece a = coverges or diverges. If i coverges, fid he i. Soluio: We calculae he i as follows ( ( ) = )( + ) + + = + = + =. + Here we divided op ad boom by =, > 0. The sequece diverges o. Quesio 5. [Sec.., #6] Deermie wheher he sequece a = coverges, fid he i. si + coverges or diverges. If i Soluio: Sice si chages sig, we look a a. Also recall ha si. Therefore 0 < si + + < for ad = 0 ( or ) + = 0. From he Squeeze Theorem, a 0 ad herefore a 0 also. Therefore he sequece coverges ad si + = 0. Quesio 6. [Sec.., #0] Deermie wheher he series e is coverge or diverge. If i is coverge, fid he sum. Soluio: Noe ha his is a geomeric series wih 0 < r = e/ < so coverges. e = (Here 0 < r = e/ <.) The sum coverges. ee = e Quesio 7. [Sec.., #] Deermie wheher he series is coverge or diverge. If i is coverge, fid he sum. ( e ) = e Soluio: This is jus he harmoic series, ( s = ) ad he series divereges. e = e e.

3 Quesio 8. [Sec.., #4] Deermie wheher he series ( + ) ( + ) is coverge or diverge. If i is coverge, fid he sum. Soluio: We check a. ( + ) ( + ) = + ( ) + ad by he Tes for Divergece he series diverges. ) = ( + ) + = = 0, Quesio 9. [Sec.., #8] Deermie wheher he series [(0.8) (0.) ] is coverge or diverge. If i is coverge, fid he sum. Soluio: We oe ha he series looks like he differece of wo geomeric series. ( ) 8 [(0.8) (0.) ] = 0 so each is a coverge geomeric series ad ( ) = 0 ( ) ( ) 0 [(0.8) (0.) ] = = = 7. Quesio 0. [Sec.., #0] Deermie wheher he series ( ) l + 5 is coverge or diverge. If i is coverge, fid he sum. Soluio: We look a he fucio l(x/(x + 5)). ( ) ( x l = l x x + 5 sice l fucio is coiuous. Hece ad he series diverges by he Tes for Divergece. x ) x = l x + 5, ( ) l = l + 5 0,

4 Quesio. [Sec.., #44] Fid he values of x for which he series (x + ) =0 coverges. Fid he sum of he series for hose values of x. Soluio: The series is a geomeric series ad i coverges iff x + < ha is, if ad oly if x + <, so ha < x + < so ha 5 < x <. The sum is s = x+ = x = x =, 5 < x <. x + Quesio. [Sec.., #0] Deermie wheher he series is coverge or diverge. Soluio: We have (.4 +. ) (.4 +. ) = Boh of hese are p-series where p =.4 > ad p =. >, respecively, hece hey boh coverge ad he sum is also coverge. Quesio. [Sec.., #6] Deermie wheher he series is coverge or diverge. + ( + ) Soluio: Le f(x) = x +, we fid he parial fracio decomposiio of he fucio firs. x(x + ) x + x(x + ) = A x + B x + ad x + = A(x + ) + Bx, herefore x + = A(x + ) + Bx so ha x = 0 implies = A ad x = implies B =. Therefore f(x) = x + x + o [, ). I is easy o see ha f is coiuous, posiive, ad decreasig sice boh /x ad /(x + ) are decreasig hece he sum is decreasig. Ad ( x + ) ( dx = x + x + ) dx = ( l x + l(x + )) x + ad he series diverges by he Iegral Tes. = ( l + l( + ) l l ) =,

5 Quesio 4. [Sec.., #4] Deermie wheher he series is coverge or diverge. = l l(l ) Soluio: Le f(x) = /x l x l(l x) o [, ), i is easy o see ha f is coiuous, posiive, ad decreasig sice he boom is icreasig, so ha dx = x l x l(l x) x l x l(l x) dx l l = du (u = l(l x), du = l l u l x x dx) l l = l u ad by he Iegral Tes, he series diverges. l l = [l(l(l )) l(l(l ))] =, Quesio 5. [Sec.., #] Fid he sum of he series correc o hree decimal places. Soluio: From he Iegral Tes we showed ha he series is coverge for p > so he above series is coverge ad he remaider erm is R R = 4 = dx = dx = x5 x5 4x 4 R = 4 6 = 64 = R = 4 4 = R 4 = = R 5 = = If we use s 5 o approximae he sum, he R , so correc o hree decimal places. = 4 4, s s 5 =

6 Quesio 6. Show ha he sequece a = q + q coverges o he same i 0 for boh q < ad q >. Soluio: Noe ha for q <, as. Also, for q >, as. 0 a = q + q < q 0 0 a = / q + / q < q 0