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Why can I use number 1 in Taylor series arctan? Taylor serie arctan: $\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$

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    $\begingroup$ Your question is unclear. $\endgroup$ – Fred Jul 16 at 17:57
  • $\begingroup$ Do you mean: why can $x$ be $1$? If so because that is within its radius of convergence. $\endgroup$ – badjohn Jul 16 at 18:00
  • $\begingroup$ @badjohn It's on the boundary. Convergence is not guaranteed. $\endgroup$ – saulspatz Jul 16 at 18:14
  • $\begingroup$ @saulspatz Good point. I guess that I should have said: it's known to converge for $1$. $\endgroup$ – badjohn Jul 16 at 18:25
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Yes, you can, because:

  • By the alternating series test, that series converges when $x=1$.
  • By Abel's theorem,$$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}=\lim_{x\to1^-}\sum_{n=0}^\infty\frac{(-1)^nx^n}{2n+1}.$$
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