TITLE: Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$
QUESTION [54 upvotes]: The equality$$\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)\tag{1}$$follows from the fact that the sum of the first series is $\dfrac\pi4$, whereas the sum of the second one is $\dfrac{\pi^2}6$.
My question is: can someone provide a proof that $(1)$ holds without using this?

REPLY [18 votes]: Here is a proof of the identity
$$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}$$
without knowing the value of either series.
Consider the double integral
\begin{align*}
I &= \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx.
\end{align*}
On one hand, we observe
\begin{align*}
 \frac{1}{(1+x^2)(1+y^2)} &= \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m, \quad |x|,|y| < 1.
\end{align*}
Then, we get that 
\begin{align*}
I &= \int_{0}^1 \int_{0}^1 \sum_{n \geq 0} \left(-x^2 \right)^n \sum_{m \geq 0} \left(-y^2 \right)^m \ dy \ dx \\
&=\sum_{n \geq 0}\sum_{m \geq 0}  \int_{0}^1 \int_{0}^1 \left(-x^2 \right)^n \left(-y^2 \right)^m \ dy \ dx \\
&=\sum_{n \geq 0}\sum_{m \geq 0} \frac{(-1)^n}{2n+1} \frac{(-1)^m}{2m+1} \\
&=\left(\sum_{n \geq 0}  \frac{(-1)^n}{2n+1} \right)^2.
\end{align*}
The interchanging of sum and integral done here can be justified by the Dominated Convergence Theorem. 
Now, we make the substitution $x \mapsto 1/x,$ and $y \mapsto 1/y$ to $I,$ resulting in
\begin{align*}
I &= \int_{1}^{\infty}\int_{1}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx,
\end{align*}
This implies
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{(1+x^2)(1+y^2)} \ dy \ dx.
\end{align*}
Make another substitution $y=tx$ with $dy = x \ dt$ to see
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dt \ dx \\
&=\frac{1}{4}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{(1+x^2)(1+t^2x^2)} \ dx \ dt, 
\end{align*}
in which the interchanging of the integral is justified by Tonelli's Theorem.
Now, using the partial fractions identity,
$$\frac{x}{(1+x^2)(1+t^2x^2)} = \frac{2 x}{\left(2-2 t^2\right) \left(x^2+1\right)}-\frac{2 t^2 x}{\left(2-2 t^2\right)
   \left(t^2 x^2+1\right)},$$
and integrating with respect to $x,$ we get
\begin{align*}
I &= \frac{1}{4}\int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt. 
\end{align*}
Now, observe 
\begin{align*}
\int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt + \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt= \int_{0}^{\infty} \frac{\log(t)}{t^2-1} \ dt.
\end{align*}
We see that 
\begin{align*}
\int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt &= \int_{1}^{\infty} \frac{\log(t)}{t^2-1} \ dt
\end{align*}
by making a change of variables $t \mapsto 1/t$ on the second integral. Thus,
\begin{align*}
I &= \frac{1}{2} \int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt.
\end{align*}
Now, convert the integrand into a geometric series
\begin{align*}
\frac{\log(t)}{t^2-1} &= - \sum_{n \geq 0} \log(t) t^{2n}, \quad t \in (0,1),
\end{align*}
and we see that
\begin{align*}
I &= -\frac{1}{2}  \int_{0}^{1}  \sum_{n \geq 0} \log(t) t^{2n} \ dt \\
&= -\frac{1}{2}  \sum_{n \geq 0} \int_{0}^{1} \log(t) t^{2n} \ dt \\
&= \frac{1}{2} \sum_{n \geq 0} \frac{1}{(2n+1)^2},
\end{align*}
in which the interchanging of sum and integral is justified by Beppo Levi's Monotone Convergence Theorem. 
Observing that 
\begin{align*}
\sum_{n \in \mathbb{N}} \frac{1}{n^2} &= \sum_{n \in \mathbb{N}} \frac{1}{(2n)^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2} \\
&=
\frac{1}{4}\sum_{n \in \mathbb{N}} \frac{1}{n^2}+\sum_{n \geq 0} \frac{1}{(2n+1)^2},
\end{align*}
we find 
\begin{align*}
I &= \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}.
\end{align*}
Hence, we have 
$$\left(\sum_{n \geq 0} \frac{(-1)^n}{(2n+1)}\right)^2 = \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}.$$
Remark
For those that may be interested, these types of integral calculations to evaluate series may be found in my joint paper with Daniele Ritelli. See https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/