How do I evaluate ∑ n = 1 ∞ ( − 1 ) n + 1 H n 2 n + 1 ?

Question

How do I evaluate       ∑          n      =      1              ∞                  (      −      1              )                  n          +          1                            H        n                    2      n      +      1        ?

Jordan Mcpherson · Accepted Answer

Using the generating function for             ∑              n        =        1                    ∞                    z              n                    H              n              =                  −        ln        ⁡        (        1        −        z        )                    1        −        z            So we have      ∑          n      =      1              ∞        (  −  1      )          n            z          n            H          n        =            −      ln      ⁡      (      1      +      z      )              1      +      z      So      ∑          n      =      1              ∞        (  −  1      )          n            z          2      n            H          n        =            −      ln      ⁡      (      1      +              z                  2                    )              1      +              z                  2                    Integrating both sides from   0  →  1 we get:-      ∑          n      =      1              ∞                  (      −      1              )                  n          +          1                            H        n                    2      n      +      1        =      ∫          0              1                  ln      ⁡      (      1      +              z                  2                    )              1      +              z                  2                        d  z  =      ∫          0                      π        4              2  ln  ⁡  (  sec  ⁡  (  x  )  )    d  x

How do I evaluate <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n

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