I would like to know if what I have written up is sufficient. I know that these are proven results (Cauchy-Hadamard), but I wanted to try my own hand at it.

Let $a_n\in\mathbb{C}\setminus\{0\}$ for $n\in\mathbb{N}_0$. Show the following:

If (1) $R=\underset{n\rightarrow\infty}{\lim}\big|\frac{a_n}{a_{n+1}}\big|$ or (2) $R=\frac{1}{\underset{n\rightarrow\infty}{\lim}\sqrt[n]{|a_n|}}$ exist, then the power series $\sum_{n=0}^{\infty}a_nz^n$ has convergence radius $R$.

(1) If we apply the root test to $a_nz^n$ we get \begin{alignat}{2} && \underset{n\rightarrow\infty}{\lim}\sqrt[n]{|a_nz^n|} &< 1\\ \Leftrightarrow &&\underset{n\rightarrow\infty}{\lim}\sqrt[n]{|a_n|} &< 1\\ \Leftrightarrow && \underset{n\rightarrow\infty}{\lim}\sqrt[n]{|a_n|}\cdot z &< 1\\ \Leftrightarrow && z &< \frac{1}{\underset{n\rightarrow\infty}{\lim}\sqrt[n]{|a_n|}} = R. \end{alignat}

which gives us the desired result. (Sorry for the misaligned equations, no idea right now how to make it pretty.)

(2) We apply the ratio test to $a_nz^n$ and get $$\underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}z^{n+1}|}{|c_nz^n|} = \underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}||z|}{|c_n|}=|z|\underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}|}{|c_n|} < 1.$$ Let $L=\underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}|}{|c_n|}$. Then $|z|\underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}|}{|c_n|}=|z|L$ and our series converges for $$|z|L<1 \Leftrightarrow |z|<\frac{1}{L} = \frac{1}{\underset{n\rightarrow\infty}{\lim}\frac{|c_{n+1}|}{|c_n|}} = |z|\underset{n\rightarrow\infty}{\lim}\frac{|c_n|}{|c_{n+1}|} = R.$$

  • $\begingroup$ Usually the root test uses $\limsup$. $\endgroup$ – Simply Beautiful Art Jan 14 '18 at 20:51
  • $\begingroup$ @SimplyBeautifulArt is that necessary here though? I assumed that just lim was fine since we assume R = lim already, hence implying absolute convergence $\endgroup$ – math_mu Jan 14 '18 at 21:00
  • $\begingroup$ Oh, okay. Should be fine then. $\endgroup$ – Simply Beautiful Art Jan 14 '18 at 22:50

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