1 1 2 + 1 1 2 + 2 2 + 1 1 2 + 2 2 + 3 2 + . . . → ? I tried like below 1 1 2 + 1 1 2 + 2 2 + 1 1 2 + 2 2 + 3 2 + ⋯ = ∑ n = 1 ∞ 1 1 2 + 2 2 + 3 2 + ⋯ + n 2 = ∑ n = 1 ∞ 1 n ( n + 1 ) ( 2 n + 1 ) 6 = ∑ n = 1 ∞ 6 n ( n + 1 ) ( 2 n + 1 ) = Then I can use the fraction ,but 1 n ( n + 1 ) ( 2 n + 1 ) = 1 n + 1 n + 1 + − 4 2 n + 1 This is ugly to turn into telescopic series . Can you help me to find :series converge to ?Thanks in advance .

Question

1          1      2        +      1                  1        2            +              2        2              +      1                  1        2            +              2        2            +              3        2              +  .  .  .  →  ?    I tried like below      1          1      2        +      1                  1        2            +              2        2              +      1                  1        2            +              2        2            +              3        2              +  ⋯  =        ∑          n      =      1              ∞            1                  1        2            +              2        2            +              3        2            +      ⋯      +              n        2              =        ∑          n      =      1              ∞            1                  n        (        n        +        1        )        (        2        n        +        1        )            6        =        ∑          n      =      1              ∞            6          n      (      n      +      1      )      (      2      n      +      1      )        =  Then I can use the fraction ,but      1          n      (      n      +      1      )      (      2      n      +      1      )        =      1    n    +      1          n      +      1        +            −      4              2      n      +      1      This is ugly to turn into telescopic series . Can you help me to find :series converge to ?Thanks in advance .

Chloe Barr · Accepted Answer

Trying to avoid Daniel Fischer&#039;s approach.      2          (      2      n      )      (      2      n      +      1      )      (      2      n      +      2      )        =      1          (      2      n      )      (      2      n      +      1      )        −      1          (      2      n      +      1      )      (      2      n      +      2      )      Collect the terms by their signs:  S  =  12      ∑          n      =      2        ∞              (      −      1              )        n                    n      (      n      +      1      )      Now apply partial fractions on this nicely and it becomes  S  =  12      ∑          n      =      2        ∞              (      −      1              )        n              n    +            (      −      1              )                  n          +          1                            n      +      1      You can then collect the terms again to get  S  =  6  +  12      ∑          n      =      3        ∞              (      −      1              )        n              n  Upon which you may use the Maclaurin expansion of   ln  ⁡  (  x  +  1  )

kakvoglq · Accepted Answer

When I post it , I find an idea ...      ∑          n      =      1              ∞            6          n      (      n      +      1      )      (      2      n      +      1      )        =        ∑          n      =      1              ∞                  6      ×      2      ×      2              (      2      n      )      (      2      n      +      2      )      (      2      n      +      1      )        =        ∑          n      =      1              ∞            24          (      2      n      )      (      2      n      +      2      )      (      2      n      +      1      )        =        ∑          n      =      1              ∞            24    1    (      1          (      2      n      )      (      2      n      +      1      )      (      2      n      +      2      )        )  =        ∑          n      =      1              ∞            24    2    (      1          (      2      n      )      (      2      n      +      1      )        −      1          (      2      n      +      1      )      (      2      n      +      2      )        )  =    12  ×      1          2      ×      3        =  2but wolfram says      ∑          n      =      1              ∞            6          n      (      n      +      1      )      (      2      n      +      1      )        =  6  (  3  −  4  l  o  g  (  2  )  )Is there something I had missed ?

1/1^2+1/(1^2+2^2)+1/(1^2+2^2+3^2)+...-> ?

Answered question

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