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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