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.
• That's a nice idea! If $(c_nr^n)$ is bounded then $\sum c_nz^n$ converges for $|z|<r$. Commented Sep 14, 2019 at 10:04
$$\{r\geq 0: c_nr^{n} \text {is bounded}\}$$ contains $$[0,R)$$ because $$r implies $$\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$$.