In my real analysis class (this is a HW question) I have been given the following problem:
- Show that $\dfrac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2 n}$ when $|x|<1$.
- Show that $\arctan (x)=\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n + 1}}{2n + 1}$ when $|x|<1$.
- 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!
Best Answer
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 $\arctan(1)=A$.