How do you find the interval of convergence Sigma n!(x-3)^n from n=[0,oo)?

1 Answer
mason m
Mar 1, 2017

The interval of convergence is x=3.

Explanation:

The series suma_n converges when L=lim_(nrarroo)abs(a_(n+1)/a_n)<1. Here, we see that

L=lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(((n+1)!(x-3)^(n+1))/(n!(x-3)^n))

Simplifying the factorial and the other part:

L=lim_(nrarroo)abs(((n+1)n!)/(n!)*(x-3)^(n+1)/(x-3)^n)

L=lim_(nrarroo)abs((n+1)(x-3))

The limit is only dependent on how n changes, so the part with x can be brought out like a constant:

L=abs(x-3)lim_(nrarroo)abs(n+1)

We see that the limit approaches oo as nrarroo. Since we want to find when L<1, as this is when the series converges, we see that the only time this will be true is when x-3=0, since then L=0.

Thus the interval of convergence is restricted to the single value x=3, and the radius of convergence is R=0.

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