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Let $(a_n)_{n \in \mathbb{N}}$ be an infinite positive sequence of integers. Assume that $f(\lambda) = \sum\limits_{n=0}^{\infty} a_n \lambda^n$ is finite for any $\lambda \in [0, \lambda_0)$. Does this necessarily imply that $f(\lambda)$ is also continuous in $(0, \lambda_0)$, or might something strange happen?

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marked as duplicate by mathcounterexamples.net, Hans Lundmark, Math1000, max_zorn, Xander Henderson Aug 4 '18 at 2:05

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    $\begingroup$ Yes, it is continuous. It is in fact analytic. $\endgroup$ – Andrés E. Caicedo Aug 3 '18 at 19:07
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    $\begingroup$ (For positive $a_n$ this is easy, because you very easily establish have uniform bounds, so you can apply Weierstrass M-test.) $\endgroup$ – Andrés E. Caicedo Aug 3 '18 at 19:09
  • $\begingroup$ Thank you. Is there a theorem I can cite without a proof being required? $\endgroup$ – QuantumLogarithm Aug 3 '18 at 19:37
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    $\begingroup$ Two results :(1). From the details of the proof of the Cauchy-Hadamard Radius Formula ( sometimes called the Hadamard Radius Formula) we can easily show that the sequence $(\sum_{n=0}^ja_n x^n)_{j\in \Bbb N}$ converges uniformly on any closed subset of $D=\{z\in \Bbb C: |z|<\lambda_0\}$....(2). A uniform limit of a sequence of continuous real or complex functions on $D$ is continuous. $\endgroup$ – DanielWainfleet Aug 3 '18 at 19:49
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If your series converges for $\lambda=r$, then it converges uniformly for all $\lambda\in(-r,r)$. This is because for such $\lambda$ $$ \sum_{n>m}a_n|\lambda|^n\leq\sum_{n>m}a_nr^n\xrightarrow[m\to\infty]{}0. $$

So $f(\lambda)$ is a uniform limit of polynomials. And a uniform limit of continuous functions is continuous.

Your function is in fact a lot more than continuous. It is analytic.

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