Radius of convergence of ∑ n = 2 ∞ n 2 n 4 n ( 2 n + 1 ) ! ( 3 − 2 x ) n

Question

Radius of convergence of      ∑          n      =      2              ∞                  n              2        n                            4        n            (      2      n      +      1      )      !        (  3  −  2  x      )    n

nelppeazy9v3ie · Accepted Answer

Factor out a -2 from   (  3  −  2  x  ). This gives      (    3    −    2    x    )    =                  (              (    −    2    )    ⋅                            −          3                2              +    (    −    2    )    ⋅    x                  )              =    (    −    2    )    (    x    −                  3        2              )    ;  so then   (  3  −  2  x      )    n    =  (  −  2      )    n    (  x  −            3      2            )    n  Recall that a power series has the form       ∑          n      =      1        ∞        c    n    (  x  −  a      )    n  ; so the       c    n   for your series is as you&#039;ve written (that your series starts at   n  =  2 doesn&#039;t matter).As for your second question, you should know that       lim          n      →      ∞                  (        1  +            1      n                          )              n    =  e. This is a fundamental limit.

Daphne Haney · Accepted Answer

I&#039;ve seen something similar around. Since you need the x part to be of the form   (  x  −  a      )    n  , you have to extract the 2 in the x:  (  3  −  2  x      )    n    =  (  −  2      )    n    ⋅  (  x  −  3      /    2      )    n  Now you have      c    n    =            (      −      2              n        2                    )        n                            4        n            (      2      n      +      1      )      !            c    n    =            (      −      2              )        n                    n                  2          n                                    4        n            (      2      n      +      1      )      !      You now need to evaluate the limit:      lim          n      →      ∞            |                  (        −        2                  )                      n            +            1                                                (            n            +            1            )                                2            (            n            +            1            )                                      (        −        2                  )          n                          n                      2            n                                                        4          n                (        2        n        +        1        )        !                              4                      n            +            1                          (        2        n        +        3        )        !              |    =      lim          n      →      ∞            |                  (        −        2        )                              (            n            +            1            )                                2            n                                                (            n            +            1            )                    2                            n                  2          n                            1              4        (        2        n        +        2        )        (        2        n        +        3        )              |    =      1    2        lim          n      →      ∞            |                                        (                                          n                +                1                            n                        )                    n                    2                                n          2                +        2        n        +        1                    4                  n          2                +        10        n        +        6              |    =      1    2        lim          n      →      ∞            |                                        (            1            +                                          1                            n                        )                    n                    2                                n          2                +        2        n        +        1                    4                  n          2                +        10        n        +        6              |    =This limits shouldn&#039;t be hard for you, I guess.            1      2              e      2              1      4        =                  e        2            8

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