Why does the Ratio Test prove absolute convergence when the limit test does not?

The question is how fast the terms tend to zero.

Consider the series

$$1+\frac12+\frac12+\frac13+\frac13+\frac13+\frac14+\frac14+\frac14+\frac14+\frac15+\frac15+\frac15+\frac15+\frac15+\cdots$$

The general term obviously tends to $0$, but if you group the terms with the same denominator, they all sum to one, making a diverging sum. This is because the decline is slow.

The ratio test is a way to qualify the decline. It ensures that the terms decrease at least as fast as

$$r^n$$ with $r<1$, and it is known that a geometric series decreases fast enough to converge, because

$$\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}\to\frac1{1-r}.$$

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In a similar vein, it is known that the generalized harmonic sequence

$$\sum_{n=0}^\infty\frac1{n^\alpha}$$ converges when $\alpha>1$. So you can base a comparison test on an estimate of $\alpha$ by computing

$$\lim_{n\to\infty}\frac{\log t_n}{\log n}.$$

Expanding a bit on Ned's comment. The explanation lies on what the ratio test actually is. If you look at a proof of it, you will see that it is basically the construction of a convergent series (one we actually know very well). So when we are using the ratio test, we are actual comparing or original series $\sum a_n$ with a convergent series $\sum b_n$.

The limit test on the other hand, is nothing of the sort. Its a observation that if a series converges it is necessary that the general term goes to zero. You are not comparing it with anything, you are just observing a fact about a series you already know converges. The question of either it is a sufficient condition is a very natural one, but it is false, as the other answers show.

The key example for the intuition is given by the harmonic series

$$\sum_{n=1}^\infty \frac1n =1+\frac12+\frac13+\frac14+\frac15+\ldots$$

for which $\frac1n\to 0$ but, using condensation idea

$$\sum_{n=1}^\infty \frac1n =1+\frac12+\overbrace{\frac13+\frac14}^{\ge 2\cdot \frac14}+\overbrace{\frac15+\frac16+\frac17+\frac18}^{\ge 4\cdot \frac18}+\ldots\ge1+\frac12+\frac12+\frac12+\ldots$$

which shows that the harmonic series diverges by comparison test.

We say that the condition $a_n\to 0$ is "only" a necessary condition for the convergence while ratio test assures a sufficient condition since it implies that the given series is dominated by a convergent geometric series.

Note that for the generalized harmonic series

$$\sum_{n=1}^\infty \frac1{n^a}, \quad\sum_{n=1}^\infty \frac1{n\log ^an} $$

ratio test fails but the trick by condensation still works and implies that both generalized harmonic series converges when $a>1$.