Math 208 - Calculus II                                                  November 13, 2006
                                    Solutions to selected problems in chapter 12
               Section 12.7 (On page 784) Test the series for convergence or divergence.
                    ∞
               27. X klnk3.
                       (k +1)
                   k=1
                                                                                                   ∞
                     Solution. Note that   klnk    ≤ klnk = lnk. Hence, by Comparison Test, X klnk
                                                 3      3        2                                           3
                                          (k +1)       k        k                                     (k +1)
                                                                 Z                             k=1
                                     ∞                             ∞                            t
                     is convergent if X lnk is convergent. But       lnk dx = lim −lnx − 1        =1. (Using
                                         k2                       2   k2       t→∞      x     x
                                     k=1                                                        1
                                                                                                     ∞
                                                                 −2                                 Xlnk
                     integration by parts with u = lnx and dv = x  ). Therefore, by the Integral test,   k2 is
                                                                                                    k=1
                                          ∞
                     convergent and so is X klnk3.
                                             (k +1)
                                         k=1
                    ∞
                   Xsin(1/n)
               33.        √n .
                   n=1
                                                                         sin(1/n)
                     Solution. UsetheL.C.T.withb =      1 . Hence, lim     √n   = lim sin(1/n) = lim sin(t) = 1.
                                                  n     3/2        n→∞     1      n→∞
                                                       n                   √              1/n      t→0   t
                                                                          n n
                           ∞                                                                ∞
                          X 1                                                   3          Xsin(1/n)
                     Since    n3/2 is convergent because it is a p−series with p = 2 > 1, then    √n    is also
                          n=1                                                              n=1
                     convergent by L.C.T.
               Chapter review exercises. (On page 823)
                                         4n
               7. Determine if {(1+3/n) } is convergent or divergent.
                                            4n         4nln(1+3/n)   limn→∞4nln(1+3/n)    12
                     Solution. lim (1+3/n)     = lim e            =e                  =e ,
                              n→∞                n→∞
                     since lim 4nln(1+3/n) = lim ln(1+3/x) = 12 (by L’Hospital). Hence, it is convergent.
                          n→∞                   x→∞       1
                                                         4x
                                      n
               8. Determine if {(−10) /n!} is convergent or divergent.
                                                                                                            n
                     Solution. Theorem 12.1.6 state that if lim |a | = 0 then lim a = 0. lim |a | = lim 10 .
                                                          n→∞ n             n→∞ n        n→∞ n       n→∞ n!
                     Since,
                                           n                             10              10
                                     0 ≤ 10 = 1010··· 1010 ··· 10 = 10     10 ··· 10 ≤ 10  10
                                          n!    1 2     1011      n    10! 11     n     10! n
                                10                                    n                                         n
                     and lim 10   10 = 0. BySqueezetheorem, lim 10 = 0. Thus,byTheorem12.1.6, lim (−10) = 0
                         n→∞ 10! n                             n→∞ n!                                 n→∞    n!
                                             ∞
                    17. Determine if X cos(3n)n is convergent or divergent.
                                                 1+(1.2)
                                            n=1
                                                                                                         ∞
                                                          cos3n             1               n           X n
                            Solution. Note the                       ≤            =(5/6) . Since             (5/6)    is a convergent GEO-
                                                                  n            n
                                                        1+(1.2)            (1.2)
                                                                                                          n=1
                                                                                            ∞                 
                                                                                           X cos(3n) 
                            METRIC series, then by the comparisons test,                                       is also convergent. Then by
                                                   ∞                                       n=11+(1.2)n
                            Theorem 12.6.3, X cos(3n)n is also convergent.
                                                       1+(1.2)
                                                  n=1
                                             ∞ √              √
                                            X n+1− n−1
                    22. Determine if                      √n             is convergent or divergent.
                                            n=1               √            √          √            √
                                                                 n+1− n−1 n+1+ n−1                                            2
                            Solution. Note that a =                    √              √            √         = √                 √          .
                                                        n                 n             n+1+ n−1 n( n+1+ n−1)
                                                                                                                                            ∞
                                                  1                                a                      2                                X 1
                            Consider b =               then show that lim            n = lim √               √       =1>0. Since,
                                          n     n3/2                        n→∞b           n→∞ x+1             x−1                              n3/2
                                                                                    n               √x + √x                                n=1
                                                                                         ∞ √              √
                                                                                       X n+1− n−1
                            is convergent (p-series), then by L.C.T. so is                            √n            .
                                                                                       n=1
                                             ∞         n√
                    26. Determine if X (−1)                n is absolutely convergent, conditionally convergent, or divergent.
                                                     lnn
                                            n=1√                 √              √                                                          √
                                                   n               x               x                                               (−1)n n
                            Solution.      lim        = lim           = lim           =∞(by L’hospital). Since lim                              6=0,
                                          n→∞lnn          x→∞lnx          n→∞ 2                                             n→∞       lnn
                            then by the Test of divergence the series is divergent.