What is the radius of convergence of #sum_1^oo (n*(x-4)^n) / (n^3 +1)#?

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Answer 1 · 2018-04-01T16:48:41

1 (The series converges when #|x-4|<1#)

The #n#-th term of the series is

#t_n = (n*(x-4)^n) / (n^3 +1)#

Then

#t_(n+1) = ((n+1)*(x-4)^(n+1)) / ((n+1)^3 +1)#

So, the ratio #t_(n+1)/t_n# is

#t_(n+1)/t_n = [((n+1)*(x-4)^(n+1)) / ((n+1)^3 +1)] div [ (n*(x-4)^n) / (n^3 +1)]#

# qquad = (n+1)/n * (n^3+1)/((n+1)^3+1) (x-4)^{n+1}/(x-4)^n#
#qquad = (1+1/n)*(1+1/n^3)/((1+1/n)^3+1)(x-4)#

Hence

#lim_{n to oo} |t_(n+1)/t_n| = lim_{n to oo} (1+1/n)*(1+1/n^3)/((1+1/n)^3+1)|x-4| = |x-4|#

Thus, according to the ratio test, the given series converges when

#|x-4|<1#