Proof for radius of convergence

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I have the following two problems related to ordinary differential equations (power series)

a. Define the radius of convergence of a series $\sum_{n=0}^{\infty}{a_{n}(x-x_{0})}^n$ for general $(a_{n})_{n\geq0}$.

b. Let $\rho$ be the radius of convergence $\sum_{n=0}^{\infty}{a_{n}(x-x_{0})}^n$. Using only the definition above, determine the radius of convergence of

$$\sum_{n=0}^{\infty}\frac{{a_{n}(x-x_{0})}^n}{n^3 \ln(n+2) e^{n}}$$

as a function of $\rho$. Justify carefully your answer.

I have solved (a) as follows

$$\rho = \sup \{r > 0: a_{n} r^n \text{ bounded} \}$$

where, $r=(x-x_{0})$

How do I solve (b)?

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As I pointed out in the comments above, by the Cauchy-Hadamard theorem (see for example [1], chapter III §2 pp. 53-54) we have the following relation $$ \rho = \sup \{r > 0: c_{n} r^n \text{ is bounded }\forall n\in\Bbb N \} \iff \rho =\frac{1}{\limsup\limits_{n\to\infty} \sqrt[n]{|c_n|}} \label{1}\tag{1} $$ Let's apply the right side formula in the relation \eqref{1} in order to calculate the convergence radius $\rho^\prime$ of the power series $$ \sum_{n=0}^{\infty}\frac{{a_{n}(x-x_{0})}^n}{n^3 \ln(n+2) e^{n}}. \label{2}\tag{2} $$ We have that $c_n=a_n\big(n^3 \ln(n+2) e^{n}\big)^{-1}$ and thus $$ \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|}= \limsup\limits_{n\to\infty} \sqrt[n]{\frac{{|a_{n}|}}{|n^3 \ln(n+2) e^{n}|}} $$ and since $\limsup\limits_{n\to\infty} \sqrt[n]{{|n^3 \ln(n+2) e^{n}|}}=e$ and $\limsup\limits_{n\to\infty} \sqrt[n]{{|a_{n}|}}=\rho^{-1}$ $$ \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|}=\frac{1}{\rho e}\iff \rho^{\prime}=\rho e $$

[1] Robert B. Burckel (1979), "An Introduction to Classical Complex Analysis", Vol. 1, Lehrbücher und Monographien aus dem Gebiete der exakten Wissenschaften. Mathematische Reihe, Band 64, Basel–Stuttgart: Birkhäuser Verlag, p. 570, ISBN 3-7643-0989-X, DOI:10.1007/978-3-0348-9374-9, MR0555733, Zbl 0434.30001.

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By ratio test we know that$$r{=\text{radius of convergence}\\=\lim_{n\to \infty}{a_{n+1}(x-x_0)^{n+1}\over a_{n}(x-x_0)^{n}}\\=(x-x_0)\cdot\lim_{n\to \infty}{a_{n+1}\over a_{n}}}$$therefore for (b) we have$$r{=(x-x_{0})\lim_{n\to \infty}\frac{{a_{n+1}}}{(n+1)^3 \ln(n+3) e^{n+1}}\cdot \frac{n^3 \ln(n+2) e^{n}}{a_n}\\={x-x_0\over e}\cdot \lim_{n\to\infty}{a_{n+1}\over a_n}}$$