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

Question

I've been taught that the ratio test proves absolute convergence when the limit test does not but I've been confused about why is that. The limit test, in my mind, gives the final value which is being continually added to the series.

For example
(Since I can't use the limit notation for the life of me, I'll type the values)
Limit of $1/n^2$, as $x$ approaches infinity, is $0$. I'd assume the sum to be something like this. $$\sum_{i=1}^\infty\left(\frac1{i^2}\right) = 1 + \ldots+ 0 + 0 + 0 + \ldots$$ So basically the value has converged and therefore there is a limit. Of course, this doesn't work for some cases like the harmonic series and I'd like to get an idea of whether thinking of them this way is correct or not and get an intuition for how to think about such series whose limit approaches $0$.

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However, There is also the Ratio Test defines absolute convergence if the limit is less than 1. That is basically telling us that the series is getting smaller and smaller and thereby telling us that the limit Limit of $f(x)$, as x approaches infinity, is 0 and basically giving the limit test. Why does this test give proof of absolute convergence then when the limit test can't give the proof?

Proof would be appreciated but intuition as to how to understand them better would be appreciated much more.

Answer 1

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

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

Answer 2

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.

Answer 3

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 $a>1$.