For what values of p is the limit test invalid? Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$. Does using the limit comparison test work when $p>1$? I'm pretty sure that the limit of $\frac{a_n}{b_n}$ diverges to $\infty$, and that $\sum b_n$ converges. So the limit comparison test doesn't tell me anything new, correct?
 A: Only if
$$0<\lim_{n\to\infty}\frac{a_n}{b_n}<\infty$$
you can conclude that  either both series converge or both series diverge. In your example the limit is equal to $0$, so the LCT is inconclusive. To show the convergence of $\sum b_n$ one could use the integral test. Note that
$$\sum_{n=1}^{\infty} {\frac{1}{n^p}}\leq \int\limits_{1}^{\infty} {\frac{1}{n^p}}\,\mathrm{d}n.$$
This integral can be easily evaluated and it can be shown that the integral, and therefore the sum, converges for $p>1$.
Addendum:
The one-sided version with the limit superior, where the zero also is included, is unfortunately inconclusive too since only from the convergence of $\sum b_n$ it follows that $\sum a_n$ converges (you want to have it the other way around).
A: Sure it works. The limit comparison test says that if $a_n, b_n \ge 0$ and
$$ \lim \frac{a_n}{b_n} = L$$
and $0 < L < \infty$ then $\sum a_n$ converges if and only if $\sum b_n$ converges.
So for instance if $a_n = \frac{1}{n(n + 1)}$ and $b_n = \frac{1}{n^2}$ then $b_n$ converges ($p > 1$) and hence so does $a_n$.
There is one special case where instead of an "if and only if" you have one-sided implication: if $L = 0$ then $\sum b_n$ converges implies $\sum a_n$ converges. I don't believe one can say anything if $L = \infty$.
For your series, $\lim a_n = 1$ (which needs to be the first thing you should notice). So if $b_n$ is any $p$-series then $\lim b_n = 0$ and hence $L = \lim a_n / b_n = \infty$. So you'll never be able to say anything using the limit comparison test for that series because $L$ will always be $\infty$. This is why you should always check if $\lim a_n = 0$ before you do anything else.
