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 $0$.
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.
Best Answer
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
$$\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}\to\frac1{1-r}.$$
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
$$\lim_{n\to\infty}\frac{\log t_n}{\log n}.$$