What theorem should I use to show that ∑ n = 0 ∞ ( − 1 ) n 4 n − 2 ( x − 2 ) ( n − 2 ) !

Question

What theorem should I use to show that      ∑          n      =      0        ∞    (  −  1      )    n                      4                  n          −          2                    (      x      −      2      )              (      n      −      2      )      !

Aryanna Caldwell · Accepted Answer

Note that   (  −  1      )    n    =  (  −  1      )          n      −      2      . Hence,      ∑          n      =      0        ∞    (  −  1      )    n                      4                  n          −          2                    (      x      −      2      )              (      n      −      2      )      !        =  (  x  −  2  )      ∑          n      =      0        ∞              (      −      4              )                  n          −          2                            (      n      −      2      )      !        =  (  x  −  2  )      (          ∑              n        =        0            ∞                      (        −        4                  )                      n                                      n        !              )  where we have interpreted             1              (        −        1        )        !              =  0  =            1              (        −        2        )        !            This is a reasonable interpretation since             1              Γ        (        0        )              =  0  =            1              Γ        (        −        1        )            Now recall that      ∑          n      =      0        ∞              y              n                    n      !        =  exp  ⁡  (  y  )  .Can you now conclude that the series converges no matter what value x takes?

Mylee Underwood · Accepted Answer

Your teacher must have given you this.      ∑          n      =      2        ∞    (  −  1      )    n                      4                  n          −          2                    (      x      −      2              )                  n          −          2                            (      n      −      2      )      !        =      ∑          n      =      0        ∞    (  −  1      )          n      +      2                          4                  n                    (      x      −      2              )        n                    n      !        =      ∑          n      =      0        ∞    (  −  1      )          n                  (      4      (      x      −      2      )              )        n                    n      !        =      e          −      4      (      x      −      2      )      for all   x  ∈      R  (by definition) which proves that regardless of the value of x the series converges.or may be this,      ∑          n      =      2        ∞    (  −  1      )    n                      4                  n          −          2                    (      x      −      2              )                  n                            (      n      −      2      )      !        =  (  x  −  2      )    2        ∑          n      =      0        ∞    (  −  1      )          n                  (      4      (      x      −      2      )              )        n                    n      !        =  (  x  −  2      )    2        e          −      4      (      x      −      2      )      for all   x  ∈      R   (by definition)or may be this,      ∑          n      =      2        ∞    (  −  1      )    n                      4                  n          −          2                    (      x      −      2      )              (      n      −      2      )      !        =  (  x  −  2  )      ∑          n      =      0        ∞    (  −  1      )          n                  4      n              n      !        =  (  x  −  2  )      e          −      4      for all   x  ∈      R   (by definition)each of which are obviously convergent for all x

What theorem should I use to show that <munderover> ∑ <mrow class="MJX-

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