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It is known that the $\displaystyle\sum_{n=0}^{\infty} a_nx^n$ converges for all $0 < x \leq M$. Let $f(x) = \displaystyle \sum_{n=0}^{\infty} a_nx^n$ for $x \in (0,M]$. Is it true that

$$\lim_{x \to 0^+} f(x) = a_0?$$

Sumanta
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Ricky The Ising
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    $$\sum_{n=0}^\infty a_nx^n\text{ converges for all }x\in (-M,M].$$ Also, $f:(-M,M]\to\Bbb R$ defined by $f(x)=\displaystyle\sum_{n=0}^\infty a_nx^n, \ x\in (-M,M]$ is continuous on $(-M,M]$. – Sumanta Jan 23 '21 at 09:32
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    Ah, thanks for the hint! Please confirm my answer:

    Since $f$ you mentioned is continuous, then $\lim_{x \to 0} f(x) = f(0)$. Then, we get what I wanted to prove.

     – Ricky The Ising Jan 23 '21 at 09:37
  • Just as information: There is no guarantee that $\displaystyle\sum_{n=0}^\infty a_n(-M)^n$ will converge, for example, $\displaystyle \sum_{n=1}^\infty\frac{(-1)^n}{n}x^n, -1<x\leq 1.$ – Sumanta Jan 23 '21 at 09:37
  • Yes, you are right. $\lim_{x\to 0+}f(x)=a_0$. – Sumanta Jan 23 '21 at 09:38
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    See also this: Abel's Limit Theorem – Sumanta Jan 23 '21 at 09:44

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