Find radius of convergence for ∑ n = 1 ∞ ( x − 1 ) n − 3 + ( x − 1 ) n − 1 4 n + 2 2 n − 1

Question

Find radius of convergence for      ∑          n      =      1              ∞                  (      x      −      1              )                  n          −          3                    +      (      x      −      1              )                  n          −          1                                    4        n            +              2                  2          n          −          1

Calvin Oneill · Accepted Answer

The way it&#039;s written, it&#039;s a weird series to ask for radius of convergence, as it has a singularity at   x  =  1 Forgetting about that, Alex Becker&#039;s answer is spot on: to find the radius of convergence you take the limit over n, not x, and only of the coefficients: here it would be  R  =      lim          n      →      ∞                  a      n              a              n        +        1              =      lim          n      →      ∞                          (                  1          4                )            n                      (                  1          4                )                    n        +        1              =  4.Or, if you want to go for understanding instead of just methods, again as Alex said this is a geometric series so it will converge when       |    x  −  1      |        /    4  &amp;lt;  1, i.e.      |    x  −  1      |    &amp;lt;  4.

Find radius of convergence for <munderover> ∑ <mrow class="MJX-TeXAtom-

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