Find the radius of convergence of $\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$ Find the radius of convergence of the following:
$$\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$$
My attempt: 
I used Ratio Test and managed to get until
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
I need to use L'Hopital Rule to get the answer. 
 A: L'Hospital's rule is not necessary here, one may just write, as $n \to \infty$,
$$
\left|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \cdot\left(\frac{n}{n+1}\right)^{3} (x-2)\right|=5\left|\frac{1 + \frac{(-1)^{n+1}}{5^{n+1}}}{1 + \frac{(-1)^{n}}{5^{n}}} \cdot\frac{1}{\left(1+\frac1n\right)^{3}} \right|\cdot|x-2| \to \color{red}{5|x-2|}.
$$
A: $$\sum_{n=1}^\infty\dfrac{5^n+(-1)^n}{n^3}(x-2)^n=\sum_{n=1}^\infty\dfrac{\{5(x-2)\}^n}{n^3}+\sum_{n=1}^\infty\dfrac{\{(-1)(x-2)\}^n}{n^3}$$
Can you apply Ratio Test separately?
OR
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
$$=\lim_{n\to\infty}\left|(x-2)\left(\dfrac1{1+\dfrac1n}\right)^3\cdot\dfrac{5-\left(\dfrac15\right)^n}{1+\left(\dfrac15\right)^n}\right|=|5(x-2)|$$
A: The root test is better here.
Taking the n-th root,
the only term that matters
is $5^n$,
with an n-th root of $5$.
Therefore,
the radius of convergence is
$\frac15$,
centered at $x=2$.
What happens at the border points
would take more precise analysis.
