The ratio test says that, for $a_k\neq 0$, if $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=L$$ exists, then if $0\leq L <1$, then $\sum_k a_k$ converges. If $L>1$, it diverges.
The notes I'm reading say that it's inadmissible to use the ratio test to test for convergence of a geometric series. I can't see why this should be the case.
Say we have some geometric series $\sum_kar^k$. Then $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{\left|ar^{k+1}\right|}{\left|ar^k\right|}=|r|.$$ So the ratio test tells us that the geometric series converges for $|r|<1$, and diverges for $|r|>1$, which is exactly what we get by using the formula $$\sum_{k=1}^n ar^k=a\left(\frac{1-r^{n+1}}{1-r}\right).$$
What is an example that demonstrates why the ratio test is inadmissible for a geometric series?
The usual proof of the ratio test is to compare the series to a geometric series. If $$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \alpha < 1,$$ then we have $$ |a_{n+1}| < |a_n| \alpha $$ for all sufficiently large $n$. It then follows from an induction argument that $$ |a_{n+k}| < |a_n| \alpha^k $$ for $n$ sufficiently large, which means that $$ \sum_{j=1}^{\infty} |a_j| = \sum_{j=1}^{n-1} |a_j| + \sum_{j=n}^{\infty} |a_j| \le \sum_{j=1}^{n-1} |a_j| + \sum_{k=0}^{\infty} |a_n| \alpha^k.$$ The first series has only finitely many terms and is therefore finite, and the second series is geometric and therefore converges by some other argument. From this, it follows that if $$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1, \qquad\text{then}\qquad \sum_{j=1}^{\infty} a_j $$ converges (by the limit comparison test, for example). The other inequality in the ratio test can be argued by noting that the general term does not go to zero, and the uncertainty at 1 can be argued by considering (for example) the harmonic and alternating harmonic series.
The key point is that we prove that the ratio test implies the convergence of series by comparison to a convergent geometric series. Since the proof of the ratio test relies on this convergence, it is circular to argue that a geometric series converges by the ratio test (unless, of course, you have another proof for the ratio test that doesn't use the convergence of geometric series).
While one could use the ratio test to establish the convergence of a geometric series (there is nothing stopping us!), it is typically poor style to rely on circular arguments as it can (potentially) lead to overlooking important hypotheses or exceptional cases. This is particularly important in a pedagogical setting, when students may not be entirely cognizant of the line of reasoning that lead up to a result (it is hard to keep track of all of the lemmata, theorems, and proofs that lead up to a result if it is the first time that you have had to deal with them).
Moreover, I don't see why one would want to use the ratio test to show that a geometric series converges. Basically no computation is needed to show that a geometric series converges, while a couple of computational steps are needed in order to invoke the ratio test.