Why is the ratio test for L=1 inconclusive?

Question

One of the often used tests for convergence (L<1) and divergence (L>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 cleary 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, no intuition for that case.

Could anybody help? Thank you!

Answer 1

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.