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How can I prove that the tailor expansion of arctan x converges only at [-1,1]? I was able to prove that it converges at the given interval but I don't know how to prove that it does not converge anywhere else . Can someone help?

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  • $\begingroup$ Well, what is the Taylor series? $\endgroup$ – zhw. Nov 29 '17 at 18:49
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    $\begingroup$ Write down the series, and find its radius of convergence. $\endgroup$ – Angina Seng Dec 3 '17 at 10:12
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The derivative of $\arctan x$ is $$\frac d{dx} \arctan x =\frac 1{1+x^2}$$

From the formula for geometric series, we have that $1+y+y^2+y^3+...=\frac 1{1−y}$ if and only if $|y|<1$.

Plugging in $-x^2$ for $y$, we get that $$\begin{align*} \frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\ &= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\ &= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots \end{align*}$$ provided that $|-x^2| \lt 1$; i.e., provided that $|x|\lt 1$.

And if $|x|\lt 1$, then $x \in [-1,1]$. If $x$ isn't in this interval, then we do not get convergence.

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    $\begingroup$ Your approach does not deal with $|x|=1$ and the Taylor series for $\arctan x$ is indeed valid for $x=\pm 1$. So you need more analysis than just term by term integration of the series for $1/(1+x^2)$. Also you need to justify why term by term integration is valid for $|x|<1$. $\endgroup$ – Paramanand Singh Nov 30 '17 at 9:42

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