Root test for radius of convergence with negative numbers

$$ S=\sum_{n=1}^{\infty}\frac{\left(-2\right)^{n}+3^{n}}{n}x^{n} $$I have to find the radius of convergence so I used the root test. Found:$$ \sqrt[n]{\left|x^{n}\right|\left|\frac{\left(-2\right)^{n}+3^{n}}{n}\right|}=\left|x\right|\sqrt[n]{\left|\left(-2\right)^{n}+3^{n}\right|} $$Not exactly sure how to find the limit inside the absolute value.
For positive numbers $a>0,b>0 $I can just use the squeeze theorem:$$ \sqrt[n]{max\left\{ a,b\right\} ^{n}}<\sqrt[n]{a^{n}+b^{n}}<\sqrt[n]{2max\left\{ a,b\right\} ^{n}} $$How do I find the limit with negative values though?

1 Answer

For each $n\in\Bbb N$,

• if $n$ is even, $(-2)^n+3^n=2^n+3^n\in\left(3^n,2\times3^n\right)$;
• if $n$ is odd, $(-2)^n+3^n=3^n-2^n\in\left(\frac123^n,3^n\right)$.

So, in both cases, you have $(-2)^n+3^n\in\left(\frac123^n,2\times3^n\right)$. And therefore$$\sqrt[n]{\left|\frac{(-2)^n+3^n}nx^n\right|}\in\left(\sqrt[n]{\frac1{2n}}3|x|,\sqrt[n]{\frac2n}3|x|\right).$$Since$$\lim_{n\to\infty}\sqrt[n]{\frac1{2n}}=1\quad\text{and}\lim_{n\to\infty}\sqrt[n]{\frac2n}=1,\quad$$you get that$$\lim_{n\to\infty}\sqrt[n]{\left|\frac{(-2)^n+3^n}nx^n\right|}=3|x|$$and that therefore the radius of convergence is $\frac13$.