TITLE: Why does this converge to $\pi/4$?
QUESTION [22 upvotes]: The infinite series...
$\pi/4 = 1 - 1/3 + 1/5 - 1/7 ...$
...is very intriguing to me and seems like a crazy coincidence (its relationship to $\pi$).  Is it actually crazy or does it have an easy to explain, logical reasoning behind it that would make it seem not so magical?

REPLY [6 votes]: The infinite series $$\pi/4 = 1-1/3+1/5-1/7+ \ ...$$ can be established by finding the expression of Taylor series
\begin{equation}
f(x) = \sum_{k=0}^\infty \frac{1}{k!} f^{(k)}(a) (x-a)^k
\end{equation}
for $\arctan(x)$ for $x \in [-1,1]$ at $a = 0$ and applying the result for $x = 1$. The finite geometric sum formula
\begin{equation}
\sum_{k=0}^n q^k = \frac{1-q^{n+1}}{1-q}, \ \ q \in \mathbb{C}, \ q \neq 1
\end{equation}
is applied to find a Taylor-form series. The uniqueness of Taylor polynomial establishes the uniqueness of Taylor series. Note that we don't hence need to calculate all derivatives of $\arctan(x)$. We calculate 
\begin{eqnarray}
\arctan(t) & = & \arctan(t) - \arctan(0) = \bigg\vert_0^t \arctan(x) = \int_0^t \frac{1}{1+x^2} dx \\
& = & \int_0^t \Big(\frac{1-(-x^2)^{n+1}}{1-(-x^2)} + \frac{(-x^2)^{n+1}}{1-(-x^2)} \Big) dx \\
& = & \int_0^t \frac{1-(-x^2)^{n+1}}{1-(-x^2)} dx + \int_0^t \frac{(-x^2)^{n+1}}{1-(-x^2)} dx \\
& = & \int_0^t \sum_{k=0}^n (-x^2)^k dx + \int_0^t \frac{(-x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-x^2)^k dx + \int_0^t \frac{(-x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t ((-1)x^2)^k dx + \int_0^t \frac{((-1)x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-1)^k (x^2)^k dx + \int_0^t \frac{(-1)^{n+1}(x^2)^{n+1}}{1+x^2} dx \\
& = & \sum_{k=0}^n \int_0^t (-1)^k x^{2k} dx + \int_0^t \frac{(-1)^{n+1}x^{2(n+1)}}{1+x^2} dx \\
& = & \sum_{k=0}^n (-1)^k \int_0^t x^{2k} dx + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+x^2} dx \\
& = & \sum_{k=0}^n (-1)^k \bigg\vert_0^t \frac{x^{2k+1}}{2k+1} + \int_0^1 \frac{(-1)^{n+1}(tx)^{2n+2}}{1+(tx)^2} t dx \\
& = & \sum_{k=0}^n (-1)^k \frac{t^{2k+1}}{2k+1} + \int_0^1 \frac{(-1)^{n+1} t^{2n+2} x^{2n+2}}{1+(tx)^2} t dx \\
& = & \sum_{k=0}^n \frac{(-1)^k}{2k+1} t^{2k+1} + \int_0^1 \frac{(-1)^{n+1} t^{2n+3} x^{2n+2}}{1+(tx)^2} dx ,
\end{eqnarray}
where $t \in \mathbb{R}$ and $n \in \mathbb{N}$. Note that $-x^2 \neq 1$ for every $x \in \mathbb{R}$. Hence we can apply the finite geometric sum formula for every $x \in \mathbb{R}$, that allows us to calculate the Taylor polynomial for every $t \in \mathbb{R}$. Assume now $t \in [-1,1]$. To obtain the limit function we calculate
\begin{eqnarray}
\Bigg| \arctan(t) & - & \sum_{k=0}^n \frac{(-1)^k}{2k+1} t^{2k+1} \Bigg| = \Bigg| \int_0^1 \frac{(-1)^{n+1}t^{2n+3}x^{2n+2}}{1+(tx)^2} dx \Bigg| \\
& \leq & \int_0^1 \Bigg| \frac{(-1)^{n+1}t^{2n+3}x^{2n+2}}{1+(tx)^2} \Bigg| dx \\
& \leq & \int_0^1 \frac{|-1|^{n+1}|t|^{2n+3}|x|^{2n+2}}{|1+(tx)^2|} dx \\
& = & \int_0^1 \frac{1^{n+1} |t|^{2n+3} x^{2n+2}}{1+(tx)^2} dx \leq \int_0^1 \frac{|t|^{2n+3} x^{2n+2}}{1} dx \\
& = & \int_0^1 |t|^{2n+3} x^{2n+2} dx = |t|^{2n+3} \int_0^1 x^{2n+2} dx \\
& = & |t|^{2n+3} \bigg\vert_0^1 \frac{1}{2n+3} x^{2n+3} = \frac{|t|^{2n+3}}{2n+3} \leq \frac{1^{2n+3}}{2n+3} \\
& = & \frac{1}{2n+3} \rightarrow 0,
\end{eqnarray}
as $n \rightarrow \infty$. Hence
\begin{eqnarray}
\arctan(x) & = & \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{(-1)^k}{2k+1} x^{2k+1} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} x^{2k+1}
\end{eqnarray}
for $x \in [-1,1]$. Now inserting $x = 1$ into the series expression of $\arctan(x)$ we obtain
\begin{eqnarray}
\pi/4 & = & \arctan(1) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} 1^{2k+1} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \\
& = & 1 - 1/3 + 1/5 - 1/7 + \ ... \ ,
\end{eqnarray}
that is the desired result. I hope that this was what you were searching for.