Radius of convergence of $ \sum^{\infty}_{n=0}\left(\frac{5^n + (-1)^n}{n^3}\right)(x-2)^n $

Question

I am trying to find the radius of convergence for the following series:

\begin{align} \sum^{\infty}_{n=0}\left(\frac{5^n + (-1)^n}{n^3}\right)(x-2)^n \end{align}

My steps are as follows:

By ratio test,

\begin{align} \left(\frac{5^{n+1} + (-1)^{n+1}}{(n+1)^3}\right)\left(\frac{n^3}{5^n + (-1)^n}\right)|x-2| =\left(\frac{5^{n+1} + (-1)^{n+1}}{5^n + (-1)^n}\right)\left(\frac{n}{n+1}\right)^3|x-2| \end{align}

As $ n $ goes to $ \infty $,

\begin{align} \left(\frac{5^{n+1} + (-1)^{n+1}}{5^n + (-1)^n}\right) \to \infty \text{ ($ 5^{n+1} $ is the dominant term)} \end{align}

\begin{align} \left(\frac{n}{n+1}\right)^3 \to 1 \text{ (by l'hopital's rule) } \end{align}

Hence, the expression becomes:

\begin{align} 1 \cdot \infty \cdot |x-2| = \infty \cdot |x-2| \text{ as $ n $ goes to $ \infty $ } \end{align}

Thus series only converges at $x=2$. This means that the radius of convergence is $0$. Somehow, the correct answer is $ \frac{1}{5} $. Could someone please advise me on why my method is wrong?

Answer 1

See that

\begin{align} \left(\frac{5^n + (-1)^n}{n^3}\right)(x-2)^n \sim {5^n(x-2)^n\over n^3} \end{align}

Thus your series converges iff $|5(x-2)| \le 1$, ie $x\in[{9\over5}, {11\over5}]$, ie $|x-2|\le{1\over5}$.

Answer 2

By the ratio test and using

$5^n+(-1)^n\sim 5^n (n\to+\infty)$,

we find

$5|x-2|$ as a limit.

$|x-2|<\frac{1}{5} \implies$ the series converge.

$|x-2|>\frac{1}{5} \implies$ it diverges.

thus, arround $x=2$, the radius convergence is $$R=\frac{1}{5}.$$