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Question: suppose we have two complex series with real coefficients, $\sum b_kz^k$ and $\sum a_k z^k$ and it is known that $\sum b_k z^k$ converges for $|z| < 1$ and diverges elsewhere. Also $\lim_{k \to \infty}\frac{a_k}{b_k} = \lambda$. Proof that $$\lim_{x \to 1-}\frac{\sum a_k z^k}{\sum b_k z^k} = \lambda $$ Hint: what do you know about $\sum_{k = 0}^{n - 1} b_kz^k$ and $\sum_{k \geq n}b_kz^k$?

My work so far: because $\sum b_k z^k$ converges on $D(0;1)$ and diverges outside of it, $\sum b_k z^k$ converges absolutely uniformly on $D(0;1)$. As $\sum b_k |z|^k$ converges, and $|z| \in \mathbb{R}$ it follows from the ratio test that $\sum a_k z^k$ must converge absolutely as well for all $|z| < 1$, so the limit makes sense and evaluates to a real number. Also $$\lim_{x \to 1-}\frac{\sum a_k x^k}{\sum b_k x^k} = \frac{\sum a_k}{\sum b_k}$$.

So I understand that the equation from the question must be true if $\frac{a_k}{b_k} = \lambda$ for all $k$, but it only approximately holds for $k$ very large. If I was paranoid and thought my teacher was lying to me I'd say that the theorem is not even true, if we choose $a_k = 1000$ for $k \leq 500$, $b_k = 1$ for $k \leq 500$, and $a_k = \lambda e^{-k^2}$, $b_k = e^{-k^2}$ for $k > 500$ the fraction won't be $\lambda$ because the upper side is $500000 + \lambda x$ and the lower side $500 + x$ where $x$ is the number $\sum_{k = 501}^\infty e^{-k^2}$ converges to.

About the hint, $\sum_{k = 0}^{n - 1} b_kz^k + \sum_{k \geq n}b_kz^k = f(z)$ for some function $z$ because $\sum b_k z^k$ converges absolutely uniformly, so it converges uniformly, but I don't see what I can do with that.

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  • $\begingroup$ Your $b_k$ series that "disproves" the claim is is not divergent at $|z|>1$ $\endgroup$ – ziggurism Nov 7 '17 at 11:43
  • $\begingroup$ I don't see how to use the hint either. But I wonder whether l'hopital can solve this problem. $\endgroup$ – ziggurism Nov 7 '17 at 15:06
  • $\begingroup$ What is meant by $x\to1^-$ ? $\endgroup$ – Yves Daoust Nov 7 '17 at 15:12
  • $\begingroup$ @YvesDaoust that $x$ approaches one from the left side, so in $\epsilon, \delta$ terms, for all $x$ that satisfy $|x - 1| < \delta$ and $x \in [0, 1[$. $\endgroup$ – Pel de Pinda Nov 7 '17 at 15:27
  • $\begingroup$ @PeldePinda: and what is $x$ ? $\endgroup$ – Yves Daoust Nov 7 '17 at 15:44

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