Why is the ratio test for $L=1$ inconclusive?

One of the often used tests for convergence ($L\lt 1$) and divergence ($L\gt 1$) of an infinite series is the ratio test.

The idea behind it, why it works is the geometric series which dominates (or not) the tested series.

My question:
With the idea in mind that the geometric series dominates (or not) the tested one, it is a mystery to me why the test is inconclusive for the case $L=1$, because the geometric series clearly diverges in the case $x\geq 1$.

I see that there are examples for cases where $L=1$ that are convergent yet, I don't get why. I have no understanding and no intuition for that case.

Could anybody help? Thank you!

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The Gauss hypergeometric series would be one example where the ratio test can give inconclusive results. –  Ｊ. Ｍ. Apr 17 '11 at 9:24
If you look into the proof of the divergence case $L \gt 1$ you'll see that $L \gt 1$ is exploited by choosing a number $r$ strictly between $1$ and $L$ in order to get a divergent minorant. This trick doesn't work if $L = 1$. –  t.b. Apr 17 '11 at 9:29
A simple example for a series for which the ratio test gives an inconclusive result is $\sum_n n^{-2}$ (or $\sum_n n^{-1}$ for that matter). –  Fabian Apr 17 '11 at 9:51

Because the ratio test doesn't involve comparison with a geometric series of ratio $L$, but rather one with ratio close to $L$.
If your series has $L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$, then for any $\epsilon >0$, $\left|\frac{a_{n+1}}{a_n}\right|$ is eventually less than $L+\epsilon$, so your series is eventually dominated by a geometric series of ratio $L+\epsilon$. If you take $\epsilon$ small enough that $L+\epsilon<1$, then this geometric series will converge, and so your original series converges.
Similarly, if $L>1$, then your series eventually dominates a geometric series of ratio $L-\epsilon > 1$, and so diverges.
But if $L=1$, neither of these approaches work. For any $\epsilon > 0$, $1+\epsilon > 1$, so the geometric series that you can show eventually dominate your series are all divergent. Similarly, $1-\epsilon<1$, so the geometric series you can show are eventually dominated by your series are all convergent.