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?
The distinction between lim and limsup
convergence-divergencelimitslimsup-and-liminfpower seriessequences-and-series
Related Solutions
You have the right ideas, but your proof could use just a little tweaking. Let $L = \limsup |a_{n+1}/a_n|$. If $L < 1$, then for $\epsilon := (1 - L)/2$, there exists $N\in \Bbb N$ such that $|a_{n+1}/a_n| < L + \epsilon$ for all $n\ge N$, i.e., $|a_{n+1}| < |a_n|(1+L)/2$ for all $n \ge N$. Thus $|a_n| \le [(1 + L)/2]^{n-N}|a_N|$ for all $n\ge N$. Since $L < 1$, $(1 + L)/2 < 1$, thus $\sum_{n = 1}^\infty [(1 + L)/2]^{n-N}$ converges. Hence, by the comparison test, $\sum_{n = 1}^\infty a_n$ converges absolutely.
In order:
If $\limsup_{n \to \infty} \sqrt[n]{|a_n|} < 1$, then even if $\lim_{n \to \infty} |a_n|$ doesn't exist, we can find a series $(b_n)$ such that $\lim_{n \to \infty} |b_n| < 1$ and $(\forall n > n_0) |a_n| < |b_n|$. For example, write $L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$, choose $\epsilon$ such that $L + \epsilon < 1$, and choose $n_0$ such that $(\forall n \geq n_0) \sqrt[n]{|a_n|} < L + \epsilon $. Then set $b_n = a_n$ for $n < n_0$ and $b_n = (L+\epsilon)^n$ for $n \geq n_0$. We can test the convergence of $(b_n)$ with the ordinary $\lim$ version of the root test, and if $(b_n)$ converges, then obviously $(a_n)$ must also converge. In this way, the $\limsup$ version of the test reduces to the $\lim$ version.
The test for convergence uses $\limsup$, whereas the test for divergence uses $\liminf$. You can find many series in which $\liminf_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and $\limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1$, some of which diverge and some of which converge. Consider $(a_n), (b_n)$ defined recursively as $a_1 = b_1 = 1$ and then $$ \begin{align*} a_n &= \begin{cases}2 a_{n-1} & \text{$n$ is a power of 3} \\ \frac{1}{2} a_{n-1} & \text{otherwise} \end{cases} \\ b_n &= \begin{cases}\frac{1}{2} b_{n-1} & \text{$n$ is a power of 3} \\ 2 a_{n-1} & \text{otherwise} \end{cases} \end{align*}$$ Then although $\limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| = \limsup_{n \to \infty} \left|\frac{b_{n+1}}{b_n} \right| = 2$ and $\liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| = \liminf_{n \to \infty} \left|\frac{b_{n+1}}{b_n} \right| = \frac{1}{2}$, $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty b_n$ diverges.
- You're right that the radius of convergence of $\sum_{n=0}^\infty n^3 z^n$ is 1, and your reasoning is correct. (On the edge of the circle of radius $1$, the series diverges.)
Best Answer
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:
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,
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 $\frac{x^2}{2} < 1$.