The distinction between lim and limsup?

Question

If you use the ratio test for a series, then you can somehow find the exact convergence radius with $\limsup \left( \frac{a_{n+1}}{a_n} \right) $, and if this equals 1, your radius of convergence is 1, while in an apparently amazingly stark contract, simply $\lim\left( \frac{a_{n+1}}{a_n} \right) $ only tells you "if" the series converges, and even that the test is inconclusive if the result is 1. How does this make sense? Why would anyone even bother having a "test" for just convergence when you can find an exact radius then?

Answer 1

I think you're confusing a few things here, such as the ratio test, and its application to power series to determine radius of convergence.

The ratio test is stated here with liminfs and limsups:

Consider an infinite series $\sum c_n$ and let \begin{align} L &= \limsup_{n\to\infty} \frac{|c_{n+1}|}{|c_n|}, \\ l &= \liminf_{n\to\infty} \frac{|c_{n+1}|}{|c_n|}. \end{align} If $L < 1$ then $\sum c_n$ converges. If $l > 1$, then $\sum c_n$ diverges.

For a general series $\sum c_n$, "radius of convergence" doesn't make sense; you really need a power series, depending on some variable $x$. The ratio test is good for determining the radius of convergence of power series. Applying the ratio test to power series,

Consider a power series $\sum a_n x^n$ and let \begin{align} L &= \limsup_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}, \\ l &= \liminf_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}. \end{align} If $R$ is the radius of convergence, then $\frac{1}{L} \le R \le \frac{1}{l}$.

Obviously, when the limit exists, then this gives you exactly the radius of convergence, but when there's a disparity between the limsup and liminf, then you get an interval of possible radii.

In particular, note that the test does not suggest that $\limsup_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = 1$ implies that the radius of $\sum a_n x^n$ is $1$, only that the radius is at least $1$. For a specific counterexample, consider the series $$\sum_{n=0}^\infty a_n x^n = 1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{4}x^4 + \frac{1}{4}x^5 + \frac{1}{8}x^6 + \frac{1}{8}x^7 + \ldots$$ Note that $\frac{|a_{n+1}|}{|a_n|} = 1$ for all even $n$, but is $\frac{1}{2}$ for odd $n$. This tells us that the radius of convergence is somewhere between $1$ and $2$. We can further see that the radius is actually $\sqrt{2}$, because $$\sum_{n=0}^\infty a_n x^n = (1 + x)\left(1 + \frac{1}{2}x^2 + \frac{1}{4}x^4 + \frac{1}{8}x^6 + \ldots\right),$$ noting that the series in the parentheses is geometric and hence converges whenever $\frac{x^2}{2} < 1$.