I know that the radius of convergence $R$ of a power series $\sum\limits_{k=0}^{\infty}a_kx^k$ is defined by $R:=\frac{1}{\limsup\limits_{k\to\infty}\sqrt[k]{|a_k|}}$ and it can be derived from the root test of infinite series.
However, I am wondering why we don't define the radius of convergence by $R:=\frac{1}{\limsup\limits_{k\to\infty}|a_k|}$?
If we assume that $a_k$ is bounded then for all $x$ with $|x|<\frac{1}{\limsup\limits_{k\to\infty}|a_k|}$ it follows that:
$
|a_kx|<\frac{|a_k|}{\limsup\limits_{k\to\infty}|a_k|}\leq 1$ such that $\sum\limits_{k=0}^{\infty}(|a_kx|)^k$ is a geomtric series. Further, $\sum\limits_{k=0}^{\infty}\sqrt[k]{|a_k|}(|a_kx|)^k$ converges and hence $\sum\limits_{k=0}^{\infty}a_kx^k$ converges as well.
Or is my reasoning flawed?
Best Answer
Simple counterexample: $a_k=2$ has radius of convergence $1$, not $1/2$.
A more extreme one: $a_k=2^k$ has radius of convergence $1/2$, not $0$.
The flaw in your reasoning is in the algebra: in order to make the connection to geometric series, you should write $\sum |a_k| |x|^k = \sum \left ( |a_k|^{1/k} |x| \right )^k$. Then if $x<\frac{1}{\limsup |a_k|^{1/k}}$ then $|a_k|^{1/k} |x|<1$ for all but finitely many $k$. On the other hand if $x>\frac{1}{\limsup |a_k|^{1/k}}$ then $|a_k|^{1/k} |x|>1$ for infinitely many $k$.