Find values of x where ∑ n = 1 ∞ ( − 1 ) n + 1 ( x − 2 ) n n 2 n

Question

Find values of x where      ∑          n      =      1              ∞                  (      −      1              )                  n          +          1                    (      x      −      2              )        n                    n              2        n

Bruce Bridges · Accepted Answer

Your       c    n   are fine. A power series has the form      ∑          n      =      1        ∞        c    n    (  x  −  a      )    n  where the coefficients       c    n   are scalars that do not depend on x, they can depend on nYour series is      ∑          n      =      1        ∞                      (        −        1                  )                      n            +            1                                      n                  2          n                      (  x  −  2      )    n  Maybe it&#039;s illustrative to write out this series more explicitly                    1        2              (    x    −    2          )      1        +                            −          1                          2          ⋅                      2            2                                (    x    −    2          )      2        +                  1                  3          ⋅                      2            3                                (    x    −    2          )      3        +    ⋯    .  The coefficients are the numbers                    1        2              ,                                 −          1                          2          ⋅                      2            2                                ,                       1                  3          ⋅                      2            3                                ,         ⋯    ;  or, in general, the coefficients are:      c    n    =                    (        −        1                  )                      n            +            1                                      n                  2          n                      .To find the radius of convergence, you can compute (note the absolute values)      lim          n      →      ∞                                    |                          c                      n            +            1                                    |                                      |                          c          n                          |                      =      lim          n      →      ∞                                    1                      (            n            +            1            )                          2                              n                +                1                                                                          1                      n                          2              n                                            =      lim          n      →      ∞                  n              (        n        +        1        )        2              =            1      2        .The radius of convergence is the reciprocal of the above limit:   1      /    (  1      /    2  )  =  2. This tells you that: the series converges whenever       |    x  −  2      |    &amp;lt;  2and the series diverges whenever       |    x  −  2      |    &amp;gt;  2When   x  =  4, the series becomes:      ∑          n      =      1        ∞                      (        −        1                  )                      n            +            1                                      n                  2          n                      (  4  −  2      )    n    =      ∑          n      =      1        ∞                      (        −        1                  )                      n            +            1                                      n                  2          n                      (  2      )    n    =      ∑          n      =      1        ∞                      (        −        1                  )                      n            +            1                              n

Find values of x where <munderover> ∑ <mrow class="MJX-TeXAtom-ORD">

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