Assignment 9 Mathematics (Model Answer) The Integral and Comparison Tests Problem: Determine converges or divergence of the series. ) (a) 0 = (b) ) (a) =8 (b) + 3) (a) = (b) 3 + ) (a) e = (b) 5) (a) =0 (b) 3 + =6 =6 = = =0 ( + ) + 3 + 3 + e + e + 5 + 6) (a) = (b) cos = e + 7) (a) + = (b) 3 + + 3 =6 5 + 3 + 3 + + 8) (a) + =8 (b) + =5 3 + 9) (a) = e (b) 3 =5 3 + 0) (a) tan = (b) tan = ) (a) tan = (b) Solution: = e sec
) a. 0 = = =. This is a P-series where P=. > so it is convergent. b. =6 P series and P = 0.5 <, so it is divergent )a. + =8 Method Use limit comparison test (l.c.t) with b = a So lim. = > 0 b + + So the two series have the same properties since is harmonic (diverges) Then also diverges + Method Use integral test f(x) = +x is positive continous & decreasing function 8 +x dx t ln + t ln0 =,So it diverges b. =6 = (+) Method =6 +6+6 Use LCT with b = P- series and K= > So it is convergent
a So lim. b +6+6 Use l hopital rule twice lim 8 3 = 0.5 > 0 +6+6 = So the two series have the same properties but is convergent Then also converges (+) Method Use integral test f(x) = (+x) is continous And decreasing function 6 (+x) dx + = +t 30 30 t,so it converges 3) a. = use basic comparison test 3 + Since 3 + > 3 < = 3 + 3 = < 3 + = Larger series is a p-series and p= so it is convergent, then the smaller series is also convergent b. + = 3 +3+ Use L.C.T with b = harmonic (Divergent) a lim +. b 3 +3+ + + 3 + 3 = > 0 3 + 3 +3+ 3
The two series have the same properties so the original series is divergent e ) a. = Use integral test f(x)= e x x this is apostive continuous decreasing function e x. x dx =- lim e x. t t x dx =- lim [e t e ]=, so the series is convergent t e b. e = +e use Basic Comparison Test +e > + e < e +e < e e = < = e then Use integral test +e f(x)= x e x is a positive continuous decreasing function x e x dx = lim t 8 t xe x dx = lim 8 t [e t e ] =, So it converges. 8e
5) a. =0 Use comparison test 3 + > 3 3 + 3 + > 3 3 + < 3 for P-series and P= 3 > As larger series converge then smaller series converges b. + 5 + Use L.C.T with b = lim 5 a +. b 5 + = which diverges because P < 0.5 [ ( +) ] 5 + [ ( +) ] ] ( +) 5 + 5 + Then two series have the same properties. It means that original series also diverges 5 + 3 + 5 + > 0 6) a. = use comparison test (cos ) (cos ) < (cos ) > = > (cos ) = = P- series P = 0 < 0 Since smaller divergent then larger diverge
b. e e+ 3 e + + < e 3 use comparison test < e 3 3 < e For e use integral test Since f(x) = e x is a continous decreasing function x Then - e x x dx = -lim t e x = e- So e is convergent Then the smaller series convergent 7) a. ++ = 5 +3 + Use L.C.T with b = harmonic (Divergent) a Then lim ++. = > 0 b 5 +3 + So the original signal is also divergent b. 3 +3+3 =6 + + Use L.C.T with b = harmonic (Divergent)
a Then lim 3 +3+3. = > 0 b + + Then two series have the same properties, So the two series are divergent 8) a. + =8 3 + Use L.C.T with b = P-series P= (Converges) lim a +. b 3 + = > 0 Then two series have the same properties, therefore, the series converges b. + 3 + Use L.C.T with b = harmonic (Divergent) lim a +. b 3 + 3 + 3 + = > 0 Then two series have same properties, So the series diverges 9) a. = e Use integral test Since f(x) =xe x f (x) = xe x + e x < 0 for x > It means that f(x) is a continuous decreasing function Then we use integral test xe x dx [( te t e t ) ( e e )] = 5, So it t e converges
b. 3 e Use integral test f(x)= x 3 e x f (x) = x 3 e x 3x e x < 0 It means that f(x) is a continuous decreasing function Then we use integral test 5 x 3 e x dx converges [( x3 3x 6x 6 t e x ) ( 53 3 5 30 6 )] = 36,So it e 5 e5 0)a. = tan Use th term test lim tan = π 0 So it diverges b. tan Use L.C.T with b = harmonic (Diverges) a lim tan. tan = π b > 0 Then two series have same properties, so the series diverges a. tan ) Use L.C.T with b = P-series with P= > (Converges)
a lim tan. = b lim tan = π > 0 Then two series have same properties, so the series converges b. = sec Use L.C.T with b = P-series with P= > (Converges) a lim b sec. sec = π > 0 Then two series have same properties, So the series is convergent Problem : Determine all values of p for which the series converges. ) =0 ( a + b ) p, a > 0, b > 0 Use L.C.T with b = p P-series (Converges) p a lim b ( a + b ) p p. Then two series have same properties So the series is convergent if p >. ( a+b )p = b p > 0 ) = p e p Case: if p > 0. Solution:
thus the series divverges. lim p e p = n Case: if p = 0. then p e p = In addition, = p e p = = it is a harmonic thus the series divverges. Case: 3 if p< 0. p e p = = = use basic comparstion test: p 0 p e p e p e p p p e p e p = ( e p ). = = =. ( e p ). this a geometric series with r = e p = <, then its converges. thus the series converges by the comarsion test. Problem 3: Estimate the error in using the indicated partial sum S n to approximate the sum of the series. () S 00, =
R 00 00 f(x)dx = 0. 0 solution: R n f(x)dx n = x dx 00 = + 00 = 5 () S 80, = + R 80 f(x)dx 80 solution: R n f(x)dx n = x + dx = tan tan 80 80 = π tan 80 < 0. 05. (3) S 00, tan = + solution: R n f(x)dx n
R 00 00 f(x)dx = tan x x + dx 00 = (tan ) (tan 00) = (π) 8 (tan 00) < 0. 0078. Problem : Determine the number of terms needed to obtain an approximation accurate to within 0 6. () = () = 5 R n = dx x R n = n n dx x 5 solution: since R n 0 6 = + n = n 0 6 = 0 6 n, 000, 000 solution: since R n 0 6 = + n = n 0 6 = 0 6 n, 000, 000 n (, 000, 000) 3 Good luc