How to appraoch solving this series?I am given the following series and asked to solve it. ∑ n = 1 ∞ ( − 2 ) n ( 2 n + 1 ) ! I recognize that this series is somewhat similar to the Taylor series for sinx which is ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! .However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the sinx series to match that, I am completely lost.

Question

How to appraoch solving this series?I am given the following series and asked to solve it.       ∑          n      =      1        ∞                      (        −        2                  )          n                            (        2        n        +        1        )        !             I recognize that this series is somewhat similar to the Taylor series for sinx which is       ∑          n      =      0        ∞    (  −  1      )    n                      x                  2          n          +          1                            (        2        n        +        1        )        !            .However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the sinx series to match that, I am completely lost.

timmeraared · Accepted Answer

Since   sin  ⁡  (  x  )  −  x  =      ∑          n      =      1        ∞    (  −  1      )    n                      x                  2          n          +          1                            (        2        n        +        1        )        !            ,                     sin        ⁡        (        x        )        −        x            x        =      ∑          n      =      1        ∞    (  −  1      )    n                      x                  2          n                            (        2        n        +        1        )        !              =      ∑          n      =      1        ∞                      (        −                  x          2                          )                      n                                      (        2        n        +        1        )        !            .Therefore, putting       x   for x,                     sin        ⁡        (                  x                )        −                  x                            x              =      ∑          n      =      1        ∞                      (        −        x                  )                      n                                      (        2        n        +        1        )        !            .And there you are

How to appraoch solving this series? I am given the following series and asked to solve it. <mu

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