Solutios to Practice Midterms Practice Midterm. a False. Couterexample: a =, b = b False. Couterexample: a =, b = c False. Couterexample: c = Y cos. = cos. + 5 = 0 sice both its exist. + 5 cos π. 5 + 5 + 5 sice cosx sice 5 + > 5 Therefore by compariso test the series cos π 5 + = coverges absolutely. 4. First, use the Ratio test to fid the radius of covergece: + x + + + + + + + + 4 x = x = x + 4 the series cov. absolutely for x < diverges for x > the radius of covergece R =. The iterval of covergece: x = + / = /, + + + = + + + coverges =0 =0 by the alt. series test.
test =0 x =, + + + = coverges by comp. + + + + + + iterval of cov. 5. False True True 4 False 5 False [, ]. + =0 =0 + = cov. p = p-series. The = Practice Midterm. a False. Couterexample: c = b False. Couterexample: a = c False. Couterexample: a = L, b =. First, fid the radius Ratio test: x + l + + l + x = l + l + l + l + = l + the series cov. absolutely whe x R =. The iterval: l + = x x lx + lx + = x < diverges whe x > x+ x+ =
x =, diverges by the itegral test. l + =0 x =, coverges, alter. series test. l + =0 The iterval is [,.. si x = x x! + x5 4. 5! +... + x si x = + x x x6! + x0 +... 5! = x x6! + x0 + + x x7 5!! + x +... 5! = x + x x6 6 +... = l coverges absolutely by compariso test, ideed: = l = = but l < / for sufficietly large = l l < for sufficietly large / The series coverges sice it is a p-series with p = / >. / 5. False False False 4 True 5 False = 6 False
7 False 8 False 9 True 0 False Practice Midterm. False. Let s + be the sum of eve terms i the partial sum s : ad s be the sum of odd terms i s : s We kow that s = s + s = s. s + = a + + a, s + + = a + + a = a + a + + a, s + = a + a + + a +. + s is positive, eootoically icreasig ad has the it Therefore s is bouded: s M for some M ad all. s ± > 0 therefore s ± M ad s ± + s±. Therefore the sequeces s ± Partial sums t = L ± = s ±. m s m = s + m= the coverge to L + L. Therefore will have o such problems o the exam.. a = + + = + + +, + s have its: a coverges. This is too difficult ad you =
+, +, therefore a as. +., = + = + + + = + + + + + + the series coverges by the compariso test with / this oe coverges, p = / >, p-series. = / 4. = = + + coverget by the alteratig series test: + + is mootoically decreasig to zero ad positive. is diverget by the compar. test: + + is diverget p-series, p = /. + + 5. By the ratio test fid the radius of covergece: x + + + x = x + = x the series cov. absolutely for x the radius of coverg. R =. < diverges for x > The iterval: x =, coverges p =, p-series = x =, coverges alt. series test = + ad = +
The iterval of covergece: [, ] Practice Midterm 4. a False, a = b False, a =.4.6. =! =!! therefore the series diverges by divergece test.. The series x coverges at x = ad diverges at x = 8. If it has fiite radius = of covergece R, the it cov. absolutely for x < R, i.e., R + < x < R + ad diverges outside of this iterval i R + 8 8 is outside of the iterval of cov. ii R + } is iside of the iterval of covergece from i: R 6 4 R 6 from ii: R 4 Therefore the series coverges for < x < 6 diverges for x > 8 ad x < 4 ad the give iformatio is sufficiet to decide what happes whe 6 x 8 ad 4 x. Therefore a c, coverges =0 b c c, ot eough iformatio =0 c 0, diverges. =0. si x = x +, dividig by x: +! =0 fx = x +!. =0
By the ratio test: + x + +! +! x = +! x +! = x + + = 0 the series coverges for all x. 4. si si = use it compariso test with ad the fact x 0 = sice both its exist = =. =. The series = si si si coverges. + 5. l + l = l l + l = = t 0 a = si si si x x = : si coverges p-series p =??? by the it comp. test the series = l + l l + t l + t = t t 0 t / + t = l Hospital rule = t 0 /t = 0 / l + l = 0. + =