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 $a_k$?
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The radius of convergence of the ratio test is greater 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}} <c $. Thus, for all $ k> p$ we would have $$ \frac{a_{p +1}}{a_{p}} <c, \quad \frac{a_{p+2}}{a_{p +1}}<c,\quad \dots,\quad \frac{a_{k}}{ a_{k-1}} <c $$ Multiplying this $ k-p $ inequalities member to member, we have $ a_k <(a_p / c^p) \cdot c^k $. Putting $N = a_p / c^p$, we assert that $$ k> p \implies a_k <N \cdot c^k. $$ Now we know that $ \lim (N)^{\frac{1}{k}} = 1 $, then $ \lim c \cdot (N)^{\frac{1}{k}} = c $. Then, from the last inequality follows $$ \limsup (a_k)^{\frac{1}{k}} \leq \limsup c \cdot (N)^{\frac {1} {k}}=c $$ This contradiction proves the fact. |
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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 $1/2$. |
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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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