Compute the series ∑ k = 0 ∞ ( 1 4 k + 1 − 1 4 k + 2 )

Question

Compute the series      ∑          k      =      0              ∞            (          1              4        k        +        1              −          1              4        k        +        2              )

aqueritztv · Accepted Answer

One way is as follows. Note that                              ∑                      k            =            0                                ∞                                    (                      1                          4              k              +              1                                −                      1                          4              k              +              2                                )                                    =                  ∑                      k            =            0                                ∞                                    ∫                      0                                1                          (                  t                      4            k                          −                  t                      4            k            +            1                          )        d        t                                          =                  ∫          0                      1                                    ∑                      k            =            0                                ∞                                    t                      4            k                          (        1        −        t        )        d        t                                          =                  ∫          0          1                (        1        −        t        )                              1                          1              −                              t                4                                                    d        t                                          =                  ∫          0          1                                                    d              t                                      (              1              +                              t                2                            )              (              1              +              t              )                                          Now use partial fractions to integrate.                              ∫          0          1                                                    d              t                                      (              1              +                              t                2                            )              (              1              +              t              )                                                          =                  1          2                          (                      ∫            0            1                                                              d                t                                            1                +                t                                              +                      ∫            0            1                                                              1                −                t                                            1                +                                  t                  2                                                              d          t          )                                                  =                              1            2                                                                          (              log              ⁡              (              1              +              t              )              +              arctan              ⁡              (              t              )              −                              1                2                            log              ⁡              (              1              +                              t                2                            )              )                        |                    0          1                                                  =                  1          2                          (          log          ⁡          (          2          )          +                      π            4                    −                      1            2                    log          ⁡          (          2          )          )                                                  =                  1          2                          (                                    log              ⁡              (              2              )                        2                    +                      π            4                    )                                                  =                  1          8                          (          π          +          log          ⁡          (          4          )          )                    Hence, we get that      ∑          k      =      0              ∞            (          1              4        k        +        1              −          1              4        k        +        2              )    =      1    8        (    π    +    log    ⁡    (    4    )    )

Compute the series <munderover> ∑ <mrow class="MJX-TeXAtom-ORD">

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