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

Question

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

Evelyn Agard · Accepted Answer

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|# Consequently, the ratio test indicates that the given series converges when #|x-4|<1#

Axel Alvarez · Answer

undefined 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.