The task is to find a sum of multiple values cos and sin to determine the value of ∑ n = 1 ∞ ( − 1 ) n n ( 2 n + 2 ) !

Question

The task is to find a sum of multiple values   cos and   sin to determine the value of      ∑          n      =      1        ∞    (  −  1      )    n        n          (      2      n      +      2      )      !

Brooks Butler · Accepted Answer

Start with the known Maclaurin series  cos  ⁡  x  =      ∑          n      =      0        ∞    (  −  1      )    n              x              2        n                    (      2      n      )      !      and  sin  ⁡  x  =      ∑          n      =      0        ∞    (  −  1      )    n              x              2        n        +        1                    (      2      n      +      1      )      !          .The given series has factorials of even numbers in the denominator, so I began by shifting indices to make them   (  2  n  )  ! to match the cosine series. After that it was mostly a matter of following my nose: at each step in the calculation below there’s really only one thing that suggests itself strongly.                              ∑                      n            =            1                    ∞                (        −        1                  )          n                          n                      (            2            n            +            2            )            !                                              =                  ∑                      n            =            2                    ∞                (        −        1                  )                      n            −            1                                                n            −            1                                (            2            n            )            !                                                            =                  ∑                      n            =            2                    ∞                (        −        1                  )                      n            −            1                                    n                      (            2            n            )            !                          −                  ∑                      n            =            2                    ∞                (        −        1                  )                      n            −            1                                    1                      (            2            n            )            !                                                            =                  ∑                      n            =            2                    ∞                (        −        1                  )                      n            −            1                                    n                      2            n            (            2            n            −            1            )            !                          +                  ∑                      n            =            2                    ∞                                      (            −            1                          )              n                                            (            2            n            )            !                                                            =                  1          2                          ∑                      n            =            2                    ∞                                      (            −            1                          )                              n                −                1                                                          (            2            n            −            1            )            !                          +                  ∑                      n            =            0                    ∞                                      (            −            1                          )              n                                            (            2            n            )            !                          −        1        +                  1          2                                                  =                  1          2                          ∑                      n            =            1                    ∞                                      (            −            1                          )              n                                            (            2            n            +            1            )            !                          +        cos        ⁡        1        −                  1          2                                                  =                  1          2                          (                      ∑                          n              =              0                        ∞                                              (              −              1                              )                n                                                    (              2              n              +              1              )              !                                −          1          )                +        cos        ⁡        1        −                  1          2                                                  =                  1          2                sin        ⁡        1        +        cos        ⁡        1        −        1                                          =                  1          2                sin        ⁡        1        +        cos        ⁡        1        −        cos        ⁡        0

shelohz0 · Accepted Answer

No idea how Wolfram does it. But I guess you just start with power series and play around.  X  =      ∑          n      =      1        ∞    (  −  1      )    n        n          (      2      n      +      2      )      !        =  −      1          4      !        +      2          6      !        −      3          8      !        +  …The annoying part is the numerators. Try doubling it so that the numerators &quot;keep up&quot; with the denominators.  2  X  =  −      2          4      !        +      4          6      !        −      6          8      !        +  …Now the numerators are consistently off the denominators by 2.  2  X  =  (  −      4          4      !        +      6          6      !        −      8          8      !        +  …  )  +  2  (      1          4      !        +      1          6      !        −      1          8      !        +  …  )That second part is pretty much   cos  ⁡  (  1  ). And the numerators of course cancel out in the first part.  2  X  =  (  −      1          3      !        +      1          5      !        −      1          7      !        +  …  )  +  2  (  cos  ⁡  (  1  )  −  1      /    2  )Now the first part is pretty much   sin  ⁡  (  1  )  2  X  =  (  sin  ⁡  (  1  )  −  1  )  +  2  cos  ⁡  (  1  )  −  1  X  =            sin      ⁡      (      1      )        2    +  cos  ⁡  (  1  )  −  1Hope that didn&#039;t seem too random. My approach is to try to build the desired series out of ones I know using various manipulations like calculus, combining and rearranging.

The task is to find a sum of multiple values cos and sin to determine the value of <mun

Answered question

Answer & Explanation

New Questions in Calculus 2