How do you find the radius of convergence #Sigma (1*4*7* * * (3n+1))/(n!)x^n# from #n=[0,oo)#?

1 Answer

The radius of convergence is #R= 1/3#

Explanation:

The ratio test states that a necessary condition for #sum_(n=0)^oo a_n# to converge is that:

#L = lim_(n->oo) abs(a_(n+1)/a_n) <= 1#

if #L<1# the condition is also sufficient and the series converges absolutely, while for #L=1# the test is inconclusive.

We then evaluate the ratio for the series at hand:

#abs(a_(n+1)/a_n) = (abs(x)^(n+1)(prod_1^(n+1) (3k+1))/((n+1)!))/(abs(x)^n(prod_1^n (3k+1))/(n!))= abs(x)(3(n+1) +1)/(n+1) = abs(x) (3n+4)/(n+1)#

so that:

#lim_(n->oo) abs(a_(n+1)/a_n) = 3abs(x)#

The series is then absolutely convergent for #absx <1/3# and divergent for #absx >1/3#with means that the radius of convergence is #R= 1/3#