I would like to show that: ∑ n = 0 ∞ 1 ( 3 n + 1 ) ( 3 n + 2 ) = π 3 3

Question

I would like to show that:      ∑          n      =      0              ∞            1          (      3      n      +      1      )      (      3      n      +      2      )        =      π          3              3

Sophia Mcdowell · Accepted Answer

Regularized the series:                              ∑                      n            =            0                    m                          1                      (            3            n            +            1            )            (            3            n            +            2            )                                              =                              ∑                      n            =            0                    m                          (                      1                          3              n              +              1                                −                      1                          3              n              +              2                                )                =                  ∑                      n            =            0                    m                          ∫          0          1                          (                      x                          3              n                                −                      x                          3              n              +              1                                )                          d                x                                          =                              ∫          0          1                          (                                    (              1              −                              x                                  3                  m                  +                  3                                            )              (              1              −              x              )                                      1              −                              x                3                                              )                          d                x        =                  ∫          0          1                                      1            −                          x                              3                m                +                3                                                          1            +            x            +                          x              2                                                d                x            Now we can take the limit by dominating convergence theorem:      ∑          n      =      0        ∞        1          (      3      n      +      1      )      (      3      n      +      2      )        =      ∫    0    1                      d            x              1      +      x      +              x        2              =                                    2                      3                          3            arctan      ⁡              (                              2            x            +            1                                3                          )            |        0    1    =      π          3              3

Kaeden Hoffman · Accepted Answer

An important trick here is that sigma and integral signs can be changed around.  ∫      ∑          n      =      a        b    f      (    n    ,    x    )      d  x  =      ∑          n      =      a        b    ∫  f      (    n    ,    x    )      d  xAnd this is because  ∫      ∑          n      =      a        b    f  (  n  ,  x  )    d  x  ∫  f      (    a    ,    x    )    +  f  (  (  a  +  1  )  ,  x  )  +  f  (  (  a  +  2  )  ,  x  )  +  ⋯  +  f      (    (    b    −    1    )    ,    x    )    +  f  (  b  ,  x  )  =  ∫  f  (  a  ,  x  )    d  x  +  ∫  f  (  (  a  +  1  )  ,  x  )    d  x  +  ⋯  +  ∫  f  (  (  b  −  1  )  ,  x  )    d  x  +  ∫  f  (  b  ,  x  )    d  xTherefore                              ∑                      n            =            0                    m                          1                      (            3            n            +            1            )            (            3            n            +            2            )                          =                                      ∑                      n            =            0                    m                          (                      1                          3              n              +              1                                −                      1                          3              n              +              2                                )                                    =                                      ∑                      n            =            0                    m                          ∫          0          1                          (                      x                          3              n                                −                      x                          3              n              +              1                                )                          d                x                            =                                      ∫          0          1                          ∑                      n            =            0                    m                          (                      x                          3              n                                −                      x                          3              n              +              1                                )                          d                x                            =                                      ∫          0          1                          (                                    (              1              −                              x                                  3                  m                  +                  3                                            )              (              1              −              x              )                                      1              −                              x                3                                              )                          d                x        =                  ∫          0          1                                      1            −                          x                              3                m                +                3                                                          1            +            x            +                          x              2                                                d                x            Also because      ∑          n      =      0        m        (          x              3        n              −          x              3        n        +        1              )    =            (      1      −              x                  3          (          m          +          1          )                    )      (      1      −      x      )              1      −              x        3            Now let us see how the final integral      ∑          n      =      0        ∞        1          (      3      n      +      1      )      (      3      n      +      2      )        =      ∫    0    1                      d            x              1      +      x      +              x        2            is evaluated.      x    2    +  x  +  1  =            (      x      +              1        2            )        2    +            (                        3                2            )        2  therefore if you skip two steps of substitution and do it once  x  +      1    2    =            3        2    t  a  n  θ  d  x  =            3        2    s  e      c          2        θ                    ∫                              d            x                                1            +            x            +                          x              2                                      =        ∫                                                            3                            2                        s            e                          c                              2                                      θ                                              3              4                        s            e                          c                              2                                      θ                                                d                    θ                                    =                              2                      3                          θ                                          =                              2                      3                          t        a                  n                      −            1                                    (                                    2              x              +              1                                      3                                )                      ⇒      ∫    0    1                      d            x              1      +      x      +              x        2              =            2              3              3        (    t    a          n              −        1              (          3              3              )    −    t    a          n              −        1              (          1              3              )    )    =            2              3              3        (          π      3        −          π      6        )    =      π          3              3

I would like to show that: <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"

Answered question

Answer & Explanation

New Questions in Calculus 2