Since $\limsup[(a_k)^{1/k}]$ is less than or equal to $\limsup[\frac{a_{k+1}}{a_k}]$ when $a_k$ is positive, how are the equations generated from the reciprocals of these expressions both the radius of convergence of a power series with coefficients $a_k$?
Two Formulas for Radius of Convergence of power series.
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0The author (Wade) says usi$n$g the Ratio Test is weaker than the Root Test when finding the radius of convergence, but I just do not see why. – 2012-02-25
3 Answers
When $a_k$ is a positive sequence, we have $ \liminf \frac{a_{n+1}}{a_n} \leq \liminf (a_k)^{1/k} \leq \limsup (a_k)^{1/k} \leq \limsup \frac{a_{n+1} }{a_n} .$
Thus, in the case that $\lim \frac{a_{n+1} }{a_n } $ exists, then so does $ \lim (a_n)^{1/n} $ and the two limits have the same value.
The Cauchy-Hadamard Theorem tells you that the radius of convergence of $\sum a_n z^n$ is $R = \frac{1}{\limsup |a_n|^{1/n} } .$ Note that $a_n \in \mathbb{C}$ is allowed, not just $a_n > 0.$
Thus, if $ \lim |a_n|^{1/n} $ exists, say it is $L_1$, then $R=1/L_1.$ And if $\lim \frac{ |a_{n+1}|}{|a_n|} $ exists, equal to $L_2$, then $L_1=L_2$ and $R = 1/L_2$ as well. Thus when the relevant limits exist, they both give you the radius of convergence.
The root test is stronger in the sense that it applies strictly more often. Whenever the ratio limit exists then so does the root limit, but not vice versa. For example, the root test tells you $\sum 0 z^n$ has infinite radius of convergence, while the ratio limit does not exist.
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0@Jim You didn't have to compensate for your downvote, now that you've told me the issue I read the question again and realized I in fact didn't answer it properly. I just wanted to know so perhaps I could edit my answer to make it better, but your answer is perfect anyway so I'll just leave this as it is. – 2012-02-25
They do not always give the same result. The radius of convergence $R$ of a power series satisfies $ 1/R \;=\; \limsup_{k\to\infty}\, (a_k)^{1/k} $ The other expression does not always give the radius of convergence. For example, if $\{a_n\}$ is the sequence $ a_n = \begin{cases}2^n & \text{if }n\text{ is even} \\ 2^{n-1} & \text{if }n\text{ is odd,}\end{cases} $ then the ratio $a_{k+1}/a_k$ alternates between $1$ and $4$. Thus $\displaystyle\limsup_{k\to\infty}\frac{a_{k+1}}{a_k} = 4$, even though the radius of convergence is $1/2$.
The radius of convergence of the ratio test is smaller than the radius of convergence of the root test. This is a consequence of the following fact: Let $ (a_k) $ be a bounded sequence of positive real numbers. It holds: $ \liminf \frac{a_{k+1}}{a_{k}} \leq \liminf (a_k)^{\frac{1}{k}}\leq\limsup (a_k)^{\frac{1}{k}} \leq \limsup \frac{a_{k +1}}{a_ {k}}\tag{*} $ In particular, if there is a limit $ \lim \frac{a_{k +1}}{a_{k}}$, there exists the limit $ \lim (a_k)^{\frac{1}{k}}$. To verify these inequalities, it suffices to prove that $ \limsup (a_k)^{\frac{1}{k}} \leq \limsup \frac{a_{k +1}}{a_{k}}$ which usually is done by contradiction. In fact, if not, there is a number $ c $ such that $\limsup\frac{a_{k +1}}{a_ {k}}\lt c \lt \limsup(a_k)^{\frac{1}{k}}$ The first of the inequalities in $(\ast)$ result in the existence of $ p \in \mathbb {N} $ such that $ k \geq p$ implies $ \frac{a_{k +1}}{a_{k}}