Consider the sum S n = ∑ r = 2 n ( − 1 ) r ( n r ) r − 1

Question

Consider the sum      S    n    =      ∑          r      =      2        n              (      −      1              )        r                                          (                                    n          r                                      )                                      r      −      1

karburitc · Accepted Answer

Note that                              S          n                                    =                  ∑                      r            =            2                    n                                                    (              −              1                              )                r                                                    r              −              1                                                                          (                                                    n              r                                                      )                                                    =        n                  ∑                      r            =            1                                n            −            1                                                              (              −              1                              )                                  r                  +                  1                                                                    r              (              r              +              1              )                                                                          (                                                                    n                −                1                            r                                                      )                                                                                      =        n                  ∑                      r            =            1                                n            −            1                                                              (              −              1                              )                                  r                  +                  1                                                      r                                                              (                                                                    n                −                1                            r                                                      )                                                    +        n                  ∑                      r            =            1                                n            −            1                                                              (              −              1                              )                r                                                    r              +              1                                                                          (                                                                    n                −                1                            r                                                      )                                                                                      =        n                  H                      n            −            1                          +        n                  ∫          0          1                                      [                          (              1              −              x                              )                                  n                  −                  1                                            −              1                        ]                    d          x                =        n                  H                      n            −            1                          −        n                              n            −            1                    n                                                  =        n                  H          n                −        n        .            Thus, we have an exact expression for       S    n   in terms of       H    n  . The asymptotic expansion follows from that of       H    n  , i.e.,      S    n    ∼  n  log  ⁡  n  −  (  1  −  γ  )  n  +      1    2    −      1          12      n        +      1          120              n        3              −  ⋯as   n  →  +  ∞

Consider the sum S n </msub> = <munderover> ∑ <mrow clas

Answered question

Answer & Explanation

New Questions in Calculus 1