Examine convergence of $\sum_{n=0}^{\infty}\frac{2+(-1)^n}{3^n}$ Using the root test on $\sum_{n=0}^{\infty}\frac{2+(-1)^n}{3^n}$ we get $\lim \sqrt[n]{\frac{2+(-1)^n}{3^n}}=\lim \frac{\sqrt[n]{2+(-1)^n}}{\sqrt[n]{3^n}}=\lim \frac{\sqrt[n]{2+(-1)^n}}{3}.$ Now I know that $\sqrt[n]{2+(-1)^n}\longrightarrow1$ and therefore we have $\frac{1}{3}<1$ and the root test shows our series convergences, but how do I show the numerator actually converges towards $1$? No matter what algebraic conversions I try to come up with, I'm not finding anything useful.
 A: $$\frac{\sqrt[n]{2+(-1)^n}}{\sqrt[n]{3^n}} \to \frac13,\quad n \to \infty.$$

how do I show the numerator actually converges towards $1$?

One may write, as $n \to \infty$,
$$
\sqrt[n]{2+(-1)^n}=e^{\large \frac1n \ln(2+(-1)^n)}\to 1
$$ since
$$
\left|\frac1n \ln(2+(-1)^n)\right| \to 0
$$ and $e^0=1$.
A: I would look at the two subsequences when $n$ is even and when $n$ is odd.  When $n$ is odd, the sequence is 
$$\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \ldots \rightarrow \frac{1}{3}.$$
When $n$ is even, the sequence is 
$$\frac{3^{1/2}}{3}, \frac{3^{1/4}}{3}, \frac{3^{1/6}}{3}, \ldots \rightarrow \frac{1}{3}.$$
Since very term appears on one of these two subsequences, we're done.
A: The root test is kind of overkill here, but it works. You're thinking too hard when it comes to the numerator - you can do it using the Squeeze Theorem!
Notice that, for any $n$, $(-1)^n$ is either $-1$ or $1$. So $2 + (-1)^n$ is either $1$ or $3$. That means that $\sqrt[n]{2 + (-1)^n}$ is either $\sqrt[n]{1}$ or $\sqrt[n]{3}$ - in particular, $\sqrt[n]{1} \leq \sqrt[n]{2 + (-1)^n} \leq \sqrt[n]{3}$. $\sqrt[n]{1} = 1$ for all $n$, so $\lim_{n \to \infty}\sqrt[n]{1} = 1$. $\lim_{n \to \infty}\sqrt[n]{3} = 1$, because $1/n$ goes to zero and $3^x$ is a continuous function. Then, by the Squeeze Theorem, $\lim_{n \to \infty}\sqrt[n]{2 + (-1)^n} = 1$.
A: the root test demands $$\lim_{n\to \infty}\sup\sqrt[n]{|a_n|}$$
A: $$\sum_{n=0}^{\infty}\frac{2+(-1)^n}{3^n}\le \sum_{n=0}^{\infty}\frac{3}{3^n} =9/2$$
