I have that $\tan^{-1}(x)=\sum_{n=1}^\infty \frac{(-1)^nx^{2n+1}}{2n+1}$ for $|x|<1$

I also know that the series $\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}$ converges since it is decreasing in absolute value and alternating.

I am wondering why we can evaluate the power series at $x=1.$ Is it because the series is a uniformly convergent sum of continuous functions defined on $x\in[0,1]$, and therefore the sum function, or $\tan^{-1}$, is continuous at $x=1$?

So for a sequence of points $\{x_n\}\in[0,1]$ with $\lim_{n\to\infty} x_n=1$, we must have that $\lim_{n\to\infty} \sum_{n=1}^\infty \frac{(-1)^n(x_n)^{2n+1}}{2n+1}= \lim_{n\to\infty} \tan^{-1}(x_n)=\tan^{-1}(1)$?

General question: if you can know that a power series of continuous functions converges on some open interval, say, $(-1,1),$ but you can show the convergence at an extremity, say $x=1,$ then is it fair to say the continuous "sum function" known to converge on $(-1,1)$ is also continuous on an interval containing an extremity, like $[0,1]$? Is continuity of the function the sole reason this is possible?


marked as duplicate by user296602, user223391, Riccardo, zz20s, Vlad Apr 17 '18 at 9:04

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ No. I was unable to follow but this might be similar? $\endgroup$ – Jungleshrimp Apr 17 '18 at 2:50

Abel's theorem states that if $\sum_{n=0}^\infty a_n$ converges, then the function defined by $f(x)=\sum_{n=0}^\infty a_n x^n$ is continuous on the half-closed interval $(-1,1]$.

In our case $a_n=0$ for even $n$, and $a_{2m+1}=(-1)^m/(2m+1)$. As $f(x)=\tan^{-1}x$ for $x\in(-1,1)$, then by continuity, $f(1)=\tan^{-1}1$.


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