How would I find the radius of convergence for ∑ n = 0 ∞ 2 − n z n 2 ? Im not sure how to deal with the z n 2 term.

Question

How would I find the radius of convergence for             ∑              n        =        0            ∞              2              −        n                    z                        n          2                    ? Im not sure how to deal with the       z                  n        2             term.

sleuteleni7 · Accepted Answer

You have the root test in your OP. The lim sup in the definition allows you to ignore the &quot;missing terms&quot; (because they are lesser) in      c    n    =      {                                        2                          −                              n                                                                                  n                    ∈                      N                                                0                                  otherwise                        and reparametrize n at will, i.e. in this case, to avoid vanishing terms; hence we take the       n    2  th rather than the nth root:  C    =        lim sup          n      →      ∞                    (              c        n            ⋅              |                    z        n                    |            )              1              /            n              =          n      →              n        2                    |    z      |    ⋅      lim          n      →      ∞                  (              1                  2          n                    )              1              /                    n        2                =        |    z      |  which converges for       |    z      |    ≤  r  =  1 since      lim          n      →      ∞            2          −      n              /                    n        2              =  1    .If you use the ratio test, you proceed with the inequality  1  &amp;gt;      lim          n      →      ∞                          2                  −          (          n          +          1          )                            z                  (          n          +          1                      )            2                                              2                  −          n                            z                              n            2                                =      lim          n      →      ∞                  z              2        n        +        1              2  to still get  r  =      lim          n      →      ∞            2          −      1              /            (      2      n      +      1      )        =  1    .

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