Page 290 - Schaum's Outline of Theory and Problems of Advanced Calculus
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CHAP. 11]                         INFINITE SERIES                               281

                                    ð M
                              If  lim  f ðxÞ dx exists and is equal to S,we see from the left-hand inequality in (1)that
                                M!1  1
                           u 2 þ u 3 þ     þ u M is monotonic increasing and bounded above by S,sothat  u n converges.
                                     M
                                    ð
                              If lim  f ðxÞ dx is unbounded, we see from the right-hand inequality in (1)that  u n diverges.
                                M!1  1
                              Thus the proof is complete.
                     11.12. Illustrate geometrically the proof in Problem
                           11.11.
                              Geometrically, u 2 þ u 3 þ     þ u M is the total area
                           of the rectangles shown shaded in Fig. 11-3, while
                           u 1 þ u 2 þ     þ u M 1 is the total area of the rectangles
                           which are shaded and nonshaded.
                              The area under the curve y ¼ f ðxÞ from x ¼ 1to
                           x ¼ M is intermediate in value between the two areas
                           given above, thus illustrating the result (1)ofProblem
                           11.11.
                                                                                   Fig. 11-3
                                                  1
                                                 X   1
                     11.13. Test for convergence:  (a)  ; p ¼ constant;
                                                    n P
                                                  1
                               1             1            1
                              X    n        X    1        X    n 2
                                      ;            ;        ne   .
                           ðbÞ    2      ðcÞ          ðdÞ
                                 n þ 1         n ln n
                               1             2             1
                                      ð M  dx  ð M    x 1 p      M  M  1 p    1
                              Consider         x  p                   where p 6¼ 1:
                                       1 x   1        1   p    1  1   p
                           ðaÞ            p  ¼    dx ¼      ¼
                                            M 1 p    1
                                 If p < 1; lim     ¼1,sothat the integral and thus the series diverges.
                                             1   p
                                        M!1
                                            M 1 p    1  1
                                 If p > 1; lim     ¼     ,sothat the integral and thus the series converges.
                                             1   p   p   1
                                        M!1
                                         ð M  dx  ð M  dx
                                  If p ¼ 1,  p  ¼   ¼ ln M and lim ln M ¼1,sothat the integral and thus the series
                              diverges.  1 x    1 x          M!1
                                  Thus, the series converges if p > 1 and diverges if p @ 1.
                                   ð M  xdx
                                                        M
                                                   2
                                                                           1
                                                                    2

                               lim        ¼ lim  1 lnðx þ 1Þj 1 ¼ lim    1  lnðM þ 1Þ  ln 2 ¼1 and the series diverges.
                                   1 x þ 1
                           ðbÞ        2         2               2          2
                               M!1          M!1            M!1
                                   M  dx
                                  ð
                                                     M
                               lim       ¼ lim lnðln xÞj 2 ¼ lim flnðln MÞ  lnðln 2Þg ¼ 1 and the series diverges.
                                   2 x ln x
                           ðcÞ
                              M!1          M!1          M!1
                                   ð  M                       n          o
                                                                    1  M
                                                   1  x M
                                                                            1  1
                               lim   xe  x 2  dx ¼ lim   e  2 j 1 ¼ lim  1  1    e  2  ¼ e  and the series converges.
                                                                e
                                                   2           2    2       2
                           ðdÞ
                               M!1  1        M!1          M!1
                                  Note that when the series converges, the value of the corresponding integral is not (in general) the
                              same as the sum of the series. However, the approximate sum of a series can often be obtained quite
                              accurately by using integrals.  See Problem 11.74.
                                       X    1     1
                                        1
                     11.14. Prove that  <       <  þ .
                                           2
                                    4     n þ 1   2  4
                                       n¼1
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