I need to determine the convergence of the series whose general term is given by: a n = 1 1 ⋅ n 2 + 1 2 ⋅ ( n − 1 ) 2 + ⋅ ⋅ ⋅ + 1 n ⋅ 1 2

Question

I need to determine the convergence of the series whose general term is given by:      a    n    =      1                  1            ⋅              n        2              +      1                  2            ⋅      (      n      −      1              )        2              +  ⋅  ⋅  ⋅  +      1                  n            ⋅              1        2

Jenna Farmer · Accepted Answer

Let   N  ∈      N  . One has                              (                      ∑                          k              =              1                        N                                1                          k                                )                          (                      ∑                          j              =              1                        N                                1                          j              2                                )                                    =                  ∑                      k            =            1                    N                          ∑                      j            =            1                    N                          1                                    j              2                                      k                                                                        ≤                  ∑                      n            =            2                                2            N                                    ∑                      j            +            k            =            n                                    1                                    j              2                                      k                                                                        =                  ∑                      n            =            1                                2            N            −            1                                    ∑                      j            +            k            =            n            +            1                                    1                                    j              2                                      k                                                                        =                  ∑                      n            =            1                                2            N            −            1                                    ∑                      k            =            1                                n                                    1                                    k                        (            n            +            1            −            k                          )              2                                          So you get that      ∑          n      =      1              2      N      −      1            a    n    ≥      (          ∑              k        =        1            N              1              k              )        (          ∑              j        =        1            N              1              j        2              )  and because             ∑              1                  k                     is divergent and             ∑              1                  j          2                     is convergent, you get that                              the series           ∑                      a            n                     diverges          .

sweetymoeyz · Accepted Answer

S    N    =      ∑          n      =      1        N        ∑          k      =      1        n        1                  k            (      n      +      1      −      k              )        2            we have  1  ≤  k  ≤  n  ≤  NSo      S    N    =      ∑          k      =      1        N        ∑          n      =      k        N        1          k            1          (      n      +      1      −      k              )        2              =      ∑          k      =      1        N        1          k            (          ∑              n        =        k            N              1              (        n        +        1        −        k                  )          2                      )  But      ∑          n      =      k        N        1          (      n      +      1      −      k              )        2              ≥      ∫          1              N      −      k      +      2            1          x      2        d  x  ≥  1  −      1          N      −      k      +      2        ≥      1    2        S    N    ≥      1    2        ∑          k      =      1        N        1          k

I need to determine the convergence of the series whose general term is given by: a n

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Answer & Explanation

New Questions in Calculus 1