Trying to show ∑ n = 1 ∞ ( 1 2 n − 1 − 1 2 ⋅ 3 n − 1 − 1 2 n + 1 ) = 1 2 log ⁡ 2

Question

Trying to show       ∑          n      =      1        ∞        (                  1                  2          n          −          1                      −                  1                  2          ⋅                      3                          n              −              1                                            −                  1                  2                      n            +            1                                )    =            1      2        log  ⁡  2

Schezzix8la5 · Accepted Answer

Rewrite   1  −            1      2        −            1      4        +            1      3        −            1      6        −            1      8        +            1      5        −  ⋯ as      (    1    −                  1        2              )    −            1      4        +      (                  1        3              −                  1        6              )    −            1      8        +      (                  1        5              −                  1        10              )    −            1      12        …or      1    2    −      1    4    +      1    6    −      1    8    +      1    10    −      1    12    +  ⋯  =      1    2        (          1      1        −          1      2        +          1      3        −          1      4        +          1      5        −          1      6        +    ⋯    )    =      1    2    log  ⁡  (  2  )  .

Trying to show <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n

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Answer & Explanation

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