# Ratio test for Series Convergence

I'm currently looking through the Wikipedia Article about the ratio test for convergence of a series. The article includes a decision diagram for the ratio test.

The diagram look something like this:

Let's take a look at $\sum_{k = 1}^{\infty} a_n$ (where $a_n \in \mathbb{R}$ or $a_n \in \mathbb{C}$ for every $n \in \mathbb{N}$).

1. $$\limsup_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| < 1$$ Means the series $\sum_{k = 1}^{\infty} a_n$ converges absolutely.
2. $$\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| > 1$$ Means the series $\sum_{k = 1}^{\infty} a_n$ diverges.
3. $$|\frac{a_{n+1}}{a_n}| \geq 1 \hspace{15px} \text{(for almost all n} \in \mathbb{N})$$ Means the series $\sum_{k = 1}^{\infty} a_n$ diverges.

Now I have two questions regarding the decision diagram.

• Can't we simply combine the requirements of the last two decisions into $$\liminf_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| \geq 1$$
• What are the requirements for the ratio test to fail (e.g. come to no conclusion)?
• The combination is only viable if you are using a subset or metric subspace I.e Cauchy sequence where $|f(x) - f(z)| < \epsilon$ for z as some parametization. The ratio test serves a purpose geometrically. A divergence could be seen as a fail that would prompt another test such as a comparison test. – McTaffy Feb 12 '17 at 18:02
• Are you sure "for almost all $n$" you wrote and the Wikipedia statement "for all large $n$" are actually the same thing? They do not seem to be to me... and that's a significant catch. You propose $\lim\inf \lvert \frac{a_{n+1}}{a_n}\rvert \geq 1$ as combination ― well, $\lim \lvert \frac{a_{n+1}}{a_n}\rvert = 1$ implies $\lim\inf \lvert \frac{a_{n+1}}{a_n}\rvert \geq 1$, and yet (quoting Wikipedia) "[if the limit is 1] the test is inconclusive, because there exist both convergent and divergent series that satisfy this case." – Clement C. Feb 12 '17 at 18:12
• @ClementC. Oh sorry. We use "almost all $n$" in my analysis lecture as "for all except finitely many $n$". So "almost all $n$" should be the same as "for all large $n$". – user326377 Feb 12 '17 at 18:20
• @Herickson OK, removed the corresponding comment. But basically, $\lim\inf_n r_n \geq 1$ and $r_n \geq 1$ for any $n$ big enough are not the same thing. You can still have $\lim\inf_n r_n \geq 1$ with infinitely many values of $r_n$ being strictly less than $1$. The Wikipedia point #3 asks that $r_n$ be actually at least $1$ for any $n$ big enough. – Clement C. Feb 12 '17 at 18:21
• @ClementC. Good point with $\lim |\frac{a_{n+1}}{a_n}| = 1 \Rightarrow \liminf |\frac{a_{n+1}}{a_n}| \geq 1$. – user326377 Feb 12 '17 at 18:22

Counter-example: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}n = \ln 2$$
But, $$\liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \liminf_{n \to \infty} \frac n{n+1} = 1$$
• Thank you. I definitely see your point. Now I'm asking myself, why not only $|\frac{a_{n+1}}{a_n}| >= 1$ for almost all $n$ is used in the decision diagram. Since ($lim_{n \rightarrow \infty } |\frac{a_{n+1}}{a_n}| > 1) \Rightarrow (|\frac{a_{n+1}}{a_n}| >= 1$ for almost all $n$). – user326377 Feb 13 '17 at 19:01
• It is true that the 2nd rule (which should be $\liminf \left|{a_{n+1}\over a_n}\right| > 1$, as can be seen in the surrounding text in the Wikipedia article) is implied by the third. However most people approach this rule as about limits, so it is common to examine it in the way shown. There are cases where the $\liminf$ is the easiest way to show that all but finitely many of the ratios is $\ge 1$. Though admittedly, that is rare. – Paul Sinclair Feb 13 '17 at 20:16