What is the reasoning behind why the ratio test works? The ratio test says that, if we have the positive series
$$\sum_{n=1}^{\infty} a_n,$$
such that $\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$, then
(1) if $L < 1$, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent;
(2) if $L > 1$, then $\sum_{n=1}^{\infty}a_n$ is divergent;
(3) if $L = 1$, then the ratio test gives no information.
I want to understand the reasons behind why and how this works, rather than just memorising formulae.
I have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \dfrac{a_{n+1}}{a_n} )$. We then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, it is more effective to take the absolute value of the limit, $\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{a_{n+1}}{a_n} }\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that, if $L < 1$, then this implies that $a_n$ has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and so as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that, if $L > 1$, then this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and so as we go to infinity, the terms diverge towards infinity.
Is this reasoning correct? If anything is incorrect, then please clarify why it is incorrect, and what the correct reasoning is.
 A: You have some ideas correct, but remember that $a_n \rightarrow 0$ does not imply that $\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct. 
For the case $L<1$, we know that there exists some $r < 1$ and $N \in \mathbb{N}$ such that $$\frac{a_{n+1}}{a_n} \leq r < 1,$$
whenever $n \geq N$. 
Therefore, 
$$
a_{N+1} \leq ra_N,
$$
$$
a_{N+2} \leq ra_{N+1}  \leq r^2a_N
$$
$$
a_{N+3} \leq ra_{N+2}  \leq r^3a_N,
$$
and in general
$$
a_{N+n} \leq r^na_N.
$$
Therefore, at least eventually, we can compare the series from above with a convergent geometric series $\sum a_N r^n$, which implies the convergence of the original series $\sum a_n$. 
A: Here is an intuitive explanation.
A geometric series has constant ratio between consecutive terms. Hence when we take the ratio $r(n)=a_{n+1}/a_n$ for an arbitrary series $\sum_k a_k,$ we are measuring how far it is similar to or different from a geometrical series.
However, this is not enough to determine the nature of the convergence or otherwise of such a series. What we need is the eventual behaviour of this series. Thus the important point is that if the ratio is constant at infinity, then the series may be called asymptotically geometric.
Hence, if the limit of the ratio $r(n)$ is less than $1$ in magnitude, the series converges. If it is greater than $1$ in magnitude, it does not converge. If the ratio at infinity is either $\pm 1,$ then this test is indecisive.
A: $L<1$ means that the ratio $r=\left|\dfrac{a_{n+1}}{a_{n}}\right|$ is eventually less than the positive ratio of a converging geometric series, i.e, less than $1$. Analogously for $L > 1$.
Note that the condition that the principal term $a_n\to 0$ when $n\to\infty$ is sufficient for a series to be convergent, but it is not a necessary condition.
A: Let's look at each of the three cases $L > 1$, $L = 1$, and $L < 1$ in detail.

*

*Case 1: Assume $L > 1$. Then there exists some $N$ so that for all $n \geq N$, we have $a_{n+1}/a_n > (L+1)/2 > 1.$ Therefore, for all $n \geq N$,

$$a_n = \frac{a_n}{a_{n-1}} \frac{a_{n-1}}{a_{n-2}}...\frac{a_{N+1}}{a_N}a_N \geq \left( \frac{L+1}{2} \right)^{n-N} a_N \geq a_N > 0,$$
so the sequence $a_n$ fails the $n$th term test: $a_n$ is bounded below by a positive number, and so the sum $\sum_{n=1}^\infty a_n$ cannot converge. (In fact, $a_n \rightarrow \infty$ when $L > 1$.)

*

*Case 2: It is possible to find both convergent and divergent series for which $L = 1$.

When $a_n = 1/n^2$, $L = 1$, and the sum $$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n^2} \text{ converges to } \frac{\pi^2}{6}.$$
When $a_n = 1/n$, $L = 1$, but the sum $$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n} \text{ diverges.}$$

*

*Case 3: Assume $L < 1$. Then there exists some $N$ so that for all $n \geq N$, we have $a_{n+1}/a_n < (L+1)/2 < 1.$ Therefore, for all $n \geq N$,

$$a_n = \frac{a_n}{a_{n-1}} \frac{a_{n-1}}{a_{n-2}}...\frac{a_{N+1}}{a_N}a_N \leq \left( \frac{L+1}{2} \right)^{n-N} a_N,$$
and we can show that $\sum_{n=1}^\infty a_n$ is bounded by the comparison test: $$\sum_{n=1}^\infty a_n = \sum_{n=1}^{N-1} a_n + \sum_{n=N}^\infty a_n \leq \sum_{n=1}^{N-1} a_n + \sum_{n=N}^\infty a_N \left( \frac{L+1}{2} \right)^{n-N};$$
but $\sum_{n=1}^{N-1} a_n$ is finite, and $$\sum_{n=N}^\infty a_N \left( \frac{L+1}{2} \right)^{n-N} = a_N \sum_{j=0}^\infty \left( \frac{L+1}{2} \right)^j$$ is a convergent geometric series (since $(L+1)/2 < 1$). QED.
A: I really though that user333870's answer covered it. But I guess I can give a short intuitive meaning.
If you apply the ratio test and discover that $L<1$, then this means that the terms of the series grow more slowly than a converging geometric series. By the Dominated Convergence Theorem that means that the series converges.
Dually if $L>1$, then this means that the terms of the series grow more quickly than the terms of a divergent geometric series, and thus diverges as well.
