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                                                 9.7 Approximation of Sums                   293

                        Proof. Equation (9.10) follows from Theorem 9.17 provided that
                                                r             1

                                                P(x) f (x) dx ≤  | f (r) − f (m)|.
                                                              2
                                              m
                        Consider the case in which f is a decreasing function; the case in which f is increasing
                        follows from a similar argument.
                          Since f is decreasing, f (x) ≤ 0on[m,r] and, since |P(x)|≤ 1/2 for all x,

                                       r                    r

                                                        1             1

                                        P(x) f (x) dx ≤−     f (x) dx =  [ f (m) − f (r)],

                                                        2             2
                                      m                   m
                        proving (9.10).
                          Equation (9.11) follows from (9.10) using the fact that

                                r           r          r          r
                                                                            1

                                   f ( j) −               f ( j) −
                                                                            2
                                            f (x) dx  ≤            f (x) dx − [ f (m) + f (r)]
                               j=m                    j=m
                                          m                     m
                                                       1
                                                     + | f (r) − f (m)|
                                                       2
                        and (9.12) follows from (9.10) by taking limits as r →∞.
                        Example 9.16 (Stirling’s approximation). Consider the function log(n!), n = 0, 1,....
                        Note that
                                                             n

                                                   log(n!) =   log( j).
                                                            j=1
                        Since log(x)isa strictly increasing function, by Corollary 9.3,
                                                n

                                     log(n!) =   log(x) dx + R n = n log(n) − n + 1 + R n
                                               1
                        where
                                                     |R n |≤ log(n)/2.
                        Hence,

                                        log(n!) = n log(n) − n + O(log(n)) as n →∞,
                        which is a crude form of Stirling’s approximation. By Example 9.8, we know that the
                        O(log(n)) term can be expanded as
                                               1         1            −1
                                                log(2π) +  log(n) + O(n ).
                                               2         2
                        Example 9.17 (Tail probability of the logarithmic series distribution). Let X denote a
                        discrete random variable such that
                                                            j
                                            Pr(X = j) = c(θ)θ /j,  j = 1, 2,...,
                        where 0 <θ < 1 and
                                                               1
                                                   c(θ) =−          .
                                                           log(1 − θ)
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