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

$|x-4|<1$

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