# Power series expansion of $\arctan (x)$ centered at $x=0$ extends to $x=1$?

In my real analysis class (this is a HW question) I have been given the following problem:

1. Show that $$\dfrac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2 n}$$ when $$|x|<1$$.
2. Show that $$\arctan (x)=\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n + 1}}{2n + 1}$$ when $$|x|<1$$.
3. Show that series for $$\arctan (x)$$ in the previous part also holds when $$x=1$$.

Here is my work for parts 1 & 2.

Since $$|x^2| < 1$$ as $$|x| < 1$$, we have that \begin{align*} \dfrac{1}{1+x^2}&= \dfrac{1}{1-(-x^2)}\\ &=\sum_{n=0}^\infty \left(-x^2\right)^n\\ &=\sum_{n=0}^\infty (-1)^n x^{2n} \end{align*} whenever $$|x| < 1$$.

We know that $$\dfrac{d}{dx}\arctan (x) =\dfrac{1}{1+x^2}$$, and thus, we may integrate termwise on the power series expansion of $$\dfrac{1}{1+x^2}$$ which we have already found: \begin{align*} \arctan(x)&= \int \sum_{n=0}^\infty (-1)^n x^{2n} dx\\ &= \sum_{n=0}^\infty \int (-1)^n x^{2n} dx\\ &= \sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{2n+1}. \end{align*}

I am unsure how to proceed with part $$3$$. I found it peculiar how this question and this question both use complex analysis to solve it, which I'm sure is not what my professor intends (we have done nothing with complex numbers in this course). If I had to guess, we are intended to use Abel's theorem, but I don't see how this question meets all the necessary requirements of Abel's Theorem. I'm looking for intuition on how to use the theorem. Thanks!

• @ClydeKertzer You can use Abel's theorem.
– Gary
Commented Oct 25, 2022 at 8:41
• @JyrkiLahtonen The question has been edited extensively since my comments, which I've deleted, were posted. Commented Oct 25, 2022 at 12:58
• I see @MarkViola. Thanks for the explanation. Everything makes sense now :-) Commented Oct 25, 2022 at 13:02
• This is famously known as the Leibniz $\pi$ formula. en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 has a neat proof without using any fancy series tools. Using the form of Abel's theorem stated here: math.stackexchange.com/questions/1827301/…, observe that $A:=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$ converges by the alternating series test, so Abel tells us $\lim_{x\nearrow 1} \arctan(x)=A$. As $\arctan(x)$ is continuous on $\mathbb R$, indeed we have $\arctan(1)=A$.
– D.R.
Commented Nov 10, 2022 at 4:40
• Thanks! If you post this as an answer I'll accept it! Commented Nov 10, 2022 at 4:50

This is famously known as the Leibniz $$\pi$$ formula. https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 has a neat proof without using any fancy series tools. Using the form of Abel's theorem stated here: Proof of Leibniz $\pi$ formula, observe that $$A:=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$ converges by the alternating series test, so Abel tells us $$\lim_{x\nearrow 1} \arctan(x)=A$$. As $$\arctan(x)$$ is continuous on $$\mathbb R$$, indeed we have $$\arctan(1)=A$$.