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

1 (The series converges when

Then

Hence

Consequently, the ratio test indicates that the given series converges when

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To find the radius of convergence of the series (\sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1}), we can use the ratio test.

Applying the ratio test, we take the limit as (n) approaches infinity of the absolute value of the ratio of consecutive terms:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)(x-4)^{n+1}}{(n+1)^3 + 1} \cdot \frac{n^3 + 1}{n(x-4)^n} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)(x-4)(n^3 + 1)}{(n+1)^3 + 1} \cdot \frac{n}{x-4} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n(n^3 + 1)(x-4)}{(n+1)^3 + 1} \right| ]

[ = |x - 4| \lim_{n \to \infty} \frac{n(n^3 + 1)}{(n+1)^3 + 1} ]

As (n) approaches infinity, the limit of this expression simplifies to (|x-4|).

For the series to converge, this limit must be less than 1. Hence, the radius of convergence is 1.

Therefore, the radius of convergence of the series is 1.

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