Clarification needed about radius of convergence

Question

My professor defined radius of convergence of power series $\sum_{n=0}^{\infty}c_nx^n$ (where $c_n\in\Bbb R$) as $R=\sup\{r\ge0:(c_nr^n )\text{ is bounded}\}$. I have not seen this kind of definition before, can someone motivate. In particular, it seems very different from ordinary definition of radius of convergence ${\displaystyle \sup \left\{|x|\ \left|\ \sum _{n=0}^{\infty }c_{n}x^{n}\ {\text{ converges }}\right.\right\}}$

He has shown that $\frac{1}{R}=\limsup_{n\to\infty}|c_n|^{\frac1n}$, but this does not clear the idea. In particular, I can't understand why boundedness of $c_nr^n$ has anything to do with convergence, I would rather think about $\sum_{k=0}^n c_nx^n$'s boundedness.

Answer 1

$\{r\geq 0: c_nr^{n} \text {is bounded}\} $ contains $[0,R)$ because $r<R$ implies $\sum c_nr^{n}$ converges so $c_nr^{n} \to 0$ making $(c_nr^{n})$ bounded. On the other hand $\{r\geq 0: c_nr^{n} \text {is bounded}\} $ cannot contain any number greater than $R$: if $R_1 >R$ and $(c_nR_1)^{n}$ is bounded then consider $\sum c_n z^{n}$ where $z=\frac {R+R_1} 2$. Since $|c_nz^{n}| =|c_nR_1^{n}| \frac {z^{n}} {R_1^{n}}$. This series is convergent by comparison with the convergent geometric series $\sum \frac {z^{n}} {R_1^{n}}$. But this contradicts the definition of the radius of convergeence (since $z >R$). It is now obvious that the supremum of $\{r\geq 0: c_nr^{n} \text {is bounded}\} $ is exactly $R$.