Question

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

Answer 1

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`.