Need help applying the root test for: $\sum\limits_{n=1}^{\infty}\left(\frac{2}{e^{-8n}-1}\right)^n$ I'm not sure if I am doing something wrong, or not... I've got an answer but it doesn't look right to me.

Given the following series, determine if it is convergent or divergent using the root or ratio test. If the test is inconclusive, use another test.
$$\sum_{n=1}^{\infty}\left(\frac{2}{e^{-8n}-1}\right)^n$$

Here's my step by step process. Maybe I left something out.
$$\lim_{n\to\infty}\left|\left(\frac{2}{e^{-8n}-1}\right)^n\right|^{\frac{1}{n}}$$
$$=\lim_{n\to\infty}\left(\frac{2}{e^{-8n}-1}\right)$$
$$=-2$$
But, even though $-2<1$, I'm very hesitant to claim that the series is convergent. I somehow feel like the answer should be positive, and I don't know if I should take the absolute value of the limit and say that the series is divergent.

You take the limit of the absolute value of the general term $\;\sqrt[\large n]{|a_n|}$, and after taking the $n$th root, what remains is still the absolute value of the nth root. Taking the limit of the absolute value of the nth root, in your case, will give you $$\lim_{n\to\infty}\left|\left(\frac{2}{e^{-8n}-1}\right)^n\right|^{\frac{1}{n}}\;=\;\;\lim_{n\to\infty}\left|\frac{2}{e^{-8n}-1}\right| = 2$$ 
Hence, in your case, the series will indeed by divergent.
