Show that $\sum_{n=0}^\infty a_n z^n$ converges $\forall z\in\mathbb{C}.$

Assume that $\sum_{n=0}^\infty b_n z^n$ converges $\forall z\in\mathbb{C}.$ Let $x=\lim_{ n\rightarrow\infty}|\frac{a_n}{b_n}|$ exists. Show that $\sum_{n=0}^\infty a_n z^n$ converges $\forall z\in\mathbb{C}.$

Proof:

Since $\sum_{n=0}^\infty b_n z^n$ converges $\forall z\in\mathbb{C},$ the radius of convergence $R_{(b_n)}=\infty, \mbox{ i.e. } \limsup_{n\rightarrow\infty} b_n = 0.$ As $x=\lim_{ n\rightarrow\infty}|\frac{a_n}{b_n}|$ exists, we can write $x=\lim_{ n\rightarrow\infty}|\frac{a_n}{b_n}|=\limsup_{n\rightarrow\infty}|\frac{a_n}{b_m}|.$ I am stuck here.

I know that to show $\sum_{n=0}^\infty a_n z^n$ converges $\forall z\in\mathbb{C},$ I need to prove that $\limsup_{n\rightarrow\infty}a_n=0.$ Any suggestions?

• But...what in the world is $\;x\;$ ?? Zero, infinity, something else...? – DonAntonio Dec 14 '13 at 19:00
• $x$ is the limit. – ugstudent1243 Dec 14 '13 at 19:02
• Er...I know that, @ugstudent1243...but if the limit is infinite then the claim may be false, for example. – DonAntonio Dec 14 '13 at 19:05
• I guess "limit exists" means by definition, limit is finite. Is it not? – ugstudent1243 Dec 14 '13 at 19:07
• Not necessarily, @ugstudent1243 . But it can be a matter of definition. – DonAntonio Dec 14 '13 at 19:10

Since the limit $\lim_{n\to \infty}\left|\frac{a_n}{b_n}\right|$ exists it follows that the sequence $\left|\frac{a_n}{b_n}\right|$ is bounded. Therefore there exists a positive number $M$ such that $\left|a_n\right|\leq M\left|b_n\right|$ for all $n\in\mathbb N$.
We conclude that $\left|a_nz^n\right|\leq M\left|b_nz^n\right|$ for all $n\in\mathbb N$ and $z\in\mathbb C$ and the result follows.

• What if $\;x=\infty\;$ ? – DonAntonio Dec 14 '13 at 19:06
• @DonAntonio The OP just clarified that $x\in\mathbb R$. But I believe it was a safe guess that he meant $x\neq\infty$ since in that case the claim is false! – P.. Dec 14 '13 at 19:10