Euler's Method for series associates with a given series ∑ j = 0 ∞ ( − 1 ) j a j the transformed series ∑ n = 0 ∞ Δ n a 0 2 n + 1 where Δ 0 a j = a j , Δ n a j = Δ n − 1 a j − Δ n − 1 a j + 1 , j = 1 , 2 , ⋯ ..I am given that ln ⁡ 2 = 1 − 1 2 + 1 3 − 1 4 + − ⋯ = 1 1 ⋅ 2 + 1 2 ⋅ 2 2 + 1 3 ⋅ 2 3 + 1 4 ⋅ 2 4 ⋯.I need to prove that ∑ n = 0 ∞ Δ n a 0 2 n + 1 = 1 2 + 1 2 ⋅ 2 2 + 1 3 ⋅ 2 3 + 1 4 ⋅ 2 4 ⋯, but I can not figure out how. I attempt to by induction and the base case is simple enough. However, when trying to use the induction hypothesis I get stuck.If I assume that ∑ n = 0 k Δ n a 0 2 n + 1 = 1 2 + 1 2 ⋅ 2 2 + 1 3 ⋅ 2 3 + 1 4 ⋅ 2 4 ⋯ 1 ( k + 1 ) ⋅ 2 k + 1 and I add 1 ( k + 2 ) ⋅ 2 k + 2 to both sides. Then I just need to show that Δ k + 1 a 0 = 1 k + 2 . However I don't see how I can.So I know that Δ k + 1 a 0 = Δ k a 0 − Δ k a 1 . From the induction hypothesis I know that Δ k a 0 = 1 k + 1 . But I would have to work with Δ k a 1 recursively. So I am not sure how to prove this.

Question

Euler&#039;s Method for series associates with a given series       ∑          j      =      0        ∞    (  −  1      )    j        a    j   the transformed series       ∑          n      =      0        ∞                      Δ        n                    a        0                    2              n        +        1             where       Δ    0        a    j    =      a    j  ,       Δ    n        a    j    =      Δ          n      −      1            a    j    −      Δ          n      −      1            a          j      +      1      ,   j  =  1  ,  2  ,  ⋯  ..I am given that   ln  ⁡  2  =  1  −      1    2    +      1    3    −      1    4    +  −  ⋯  =      1          1      ⋅      2        +      1          2      ⋅              2        2              +      1          3      ⋅              2        3              +      1          4      ⋅              2        4              ⋯.I need to prove that       ∑          n      =      0        ∞                      Δ        n                    a        0                    2              n        +        1              =      1    2    +      1          2      ⋅              2        2              +      1          3      ⋅              2        3              +      1          4      ⋅              2        4              ⋯, but I can not figure out how. I attempt to by induction and the base case is simple enough. However, when trying to use the induction hypothesis I get stuck.If I assume that       ∑          n      =      0        k                      Δ        n                    a        0                    2              n        +        1              =      1    2    +      1          2      ⋅              2        2              +      1          3      ⋅              2        3              +      1          4      ⋅              2        4              ⋯      1          (      k      +      1      )      ⋅              2                  k          +          1                     and I add       1          (      k      +      2      )      ⋅              2                  k          +          2                     to both sides. Then I just need to show that       Δ          k      +      1            a    0    =      1          k      +      2      . However I don&#039;t see how I can.So I know that       Δ          k      +      1            a    0    =      Δ          k            a    0    −      Δ          k            a    1  . From the induction hypothesis I know that       Δ          k            a    0    =      1          k      +      1      . But I would have to work with       Δ          k            a    1   recursively. So I am not sure how to prove this.

Misael Li · Accepted Answer

A formula for       Δ    n        a    j   is given by                              Δ          n                          a          j                =                              n            !                         j            !                                (            n            +            j            +            1            )            !                              We now need to use double induction ... it suffice to show that the formula       Δ    n        a    j    =      Δ          n      −      1            a    j    −      Δ          n      −      1            a          j      +      1       holds, which is easy                              Δ          n                          a          j                =                  Δ                      n            −            1                                    a          j                −                  Δ                      n            −            1                                    a                      j            +            1                          =                              (            n            −            1            )            !                         j            !                                (            n            +            j            )            !                          −                              (            n            −            1            )            !                         (            j            +            1            )            !                                (            n            +            j            +            1            )            !                          =                              n            !                         j            !                                (            n            +            j            +            1            )            !

Euler's Method for series associates with a given series <munderover> <mo movablelimits="fa

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