We would like to prove that

π=41−43+45−47+49⋯ \begin{align} \pi = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} \cdots \end{align}

This is a corollary of the Leibniz's Formula

Theorem.

arctan⁡(x)≐∑n=0∞(−1)nx2n+12n+1,∀x∈[0,1]. \arctan(x) \doteq \sum_{n = 0}^\infty (-1)^n \frac{x^{2n + 1}}{2n + 1}, \quad \forall x \in [0, 1].

Substituting x=1x = 1 establishes formula (1)(1).

Proof. First, observe that

arctan⁡′(t)=11+t2=∑n=0∞(−1)nt2n,∀t∈[0,1). \arctan'(t) = \frac{1}{1 + t^2} = \sum_{n = 0}^\infty (-1)^n t^{2n}, \quad \forall t \in [0, 1).

For each x∈[0,1)x \in [0, 1), Since the sequence ∑n=0∞(−1)nt2n\sum_{n = 0}^\infty (-1)^n t^{2n} converges uniformly for t∈[0,x)t \in [0, x),

arctan⁡(x)=∫0xarctan⁡′(t) dt=∫0x∑n=0∞(−1)nt2n dt=∑n=0∞∫0x(−1)nt2n dt=∑n=0∞(−1)nx2n+12n+1. \begin{aligned} \arctan(x) &= \int_0^x \arctan'(t) \, dt \\ &= \int_0^x \sum_{n = 0}^\infty (-1)^n t^{2n} \, dt \\ &= \sum_{n = 0}^\infty \int_0^x (-1)^n t^{2n} \, dt \\ &= \sum_{n = 0}^\infty (-1)^n \frac{x^{2n + 1}}{2n + 1}. \end{aligned}

It is left to prove that the last equality also holds for x=1x = 1. This is a result of the following lemma, which is a special case of the Abel's test.

Lemma. Let

f(x)≐∑n=0∞(−1)nx2n+12n+1,∀x∈[0,1]. f(x) \doteq \sum_{n = 0}^\infty (-1)^n \frac{x^{2n + 1}}{2n + 1}, \quad \forall x \in [0, 1].

Then f(1)f(1) is well-defined, and is left-continuous at x=1x = 1.

Proof 1 of Lemma. Note that

A≐f(1)=∑n=0∞(−1)n12n+1. A \doteq f(1) = \sum_{n = 0}^\infty (-1)^n \frac{1}{2n + 1}.

It is easy to see that the sequence converges, and hence f(1)f(1) is well-defined.

Define

sn=∑ℓ=0n(−1)ℓ12ℓ+1,∀n∈N. s_n = \sum_{\ell = 0}^n (-1)^\ell \frac{1}{2\ell + 1}, \quad \forall n \in \mathbb{N}.

Then for integers n,m∈N+n, m \in \mathbb{N}^+ s.t., n≥m+1n \ge m + 1, we have

fn(x)−fm+1(x)=∑ℓ=m+1n(sℓ−sℓ−1)x2ℓ+1=snx2n+1−smx2m+3+∑ℓ=m+1n−1sℓ(x2ℓ+1−x2ℓ+3)=snx2n+1−smx2m+3+(1−x2)∑ℓ=m+1n−1sℓx2ℓ+1. \begin{aligned} f_n(x) - f_{m + 1} (x) &= \sum_{\ell = m + 1}^n (s_{\ell} - s_{\ell - 1}) x^{2\ell + 1} \\ &= s_n x^{2n + 1} - s_{m} x^{2m + 3} + \sum_{\ell = m + 1}^{n - 1} s_{\ell} (x^{2\ell + 1} - x^{2\ell + 3}) \\ &= s_n x^{2n + 1} - s_{m} x^{2m + 3} + (1 - x^2) \sum_{\ell = m + 1}^{n - 1} s_{\ell} x^{2\ell + 1}. \end{aligned}

Given ϵ>0\epsilon > 0, ∃N∈N\exists N \in \mathbb{N}, such that $ |s_\ell - A| \le \epsilon, \forall \ell \ge N. $ Hence,

snx2n+1−smx2m+3+(1−x2)∑ℓ=m+1n−1sℓx2ℓ+1≤(A+ϵ)x2n+1−(A−ϵ)x2m+3+(1−x2)∑ℓ=m+1n−1(A+ϵ)x2ℓ+1≤(A+ϵ)x2n+1−(A−ϵ)x2m+3+(1−x2)⋅(A+ϵ)⋅x2m+3−x2n+11−x2=2⋅ϵ⋅x2m+3. \begin{aligned} &s_n x^{2n + 1} - s_{m} x^{2m + 3} + (1 - x^2) \sum_{\ell = m + 1}^{n - 1} s_{\ell} x^{2\ell + 1} \\ \le &(A + \epsilon) x^{2n + 1} - (A - \epsilon) x^{2m + 3} + (1 - x^2) \sum_{\ell = m + 1}^{n - 1} (A + \epsilon) x^{2\ell + 1} \\ \le &(A + \epsilon) x^{2n + 1} - (A - \epsilon) x^{2m + 3} + (1 - x^2) \cdot (A + \epsilon) \cdot \frac{x^{2m + 3} - x^{2n + 1}}{1 - x^2} \\ = & 2 \cdot \epsilon \cdot x^{2m + 3}. \end{aligned}

On the other hand,

snx2n+1−smx2m+3+(1−x2)∑ℓ=m+1n−1sℓx2ℓ+1≥(A−ϵ)x2n+1−(A+ϵ)x2m+3+(1−x2)∑ℓ=m+1n−1(A−ϵ)x2ℓ+1≤(A−ϵ)x2n+1−(A+ϵ)x2m+3+(1−x2)⋅(A−ϵ)⋅x2m+3−x2n+11−x2=−2⋅ϵ⋅x2m+3. \begin{aligned} &s_n x^{2n + 1} - s_{m} x^{2m + 3} + (1 - x^2) \sum_{\ell = m + 1}^{n - 1} s_{\ell} x^{2\ell + 1} \\ \ge &(A - \epsilon) x^{2n + 1} - (A + \epsilon) x^{2m + 3} + (1 - x^2) \sum_{\ell = m + 1}^{n - 1} (A - \epsilon) x^{2\ell + 1} \\ \le &(A - \epsilon) x^{2n + 1} - (A + \epsilon) x^{2m + 3} + (1 - x^2) \cdot (A - \epsilon) \cdot \frac{x^{2m + 3} - x^{2n + 1}}{1 - x^2} \\ = & - 2 \cdot \epsilon \cdot x^{2m + 3}. \end{aligned}

Therefore,

∣fn(x)−fm+1(x)∣≤2⋅ϵ. |f_n(x) - f_{m + 1} (x)| \le 2 \cdot \epsilon.

□\square

Proof 2 of Lemma. Note that

A≐f(1)=∑n=0∞(−1)n12n+1. A \doteq f(1) = \sum_{n = 0}^\infty (-1)^n \frac{1}{2n + 1}.

It is easy to see that the sequence converges, and hence f(1)f(1) is well-defined.

Define

sn=∑ℓ=0n(−1)ℓ12ℓ+1,∀n∈N. s_n = \sum_{\ell = 0}^n (-1)^\ell \frac{1}{2\ell + 1}, \quad \forall n \in \mathbb{N}.

Since lim⁡n→∞∣sn∣n=1\lim_{n \rightarrow \infty} \sqrt[n]{|s_n|} = 1, the following series converge

h(x)≐∑n=0∞snx2n+1,x∈[0,1). h(x) \doteq \sum_{n = 0}^\infty s_n x^{2n + 1}, \quad x \in [0, 1).

Hence, for each x∈[0,1)x \in [0, 1),

h(x)−x2⋅h(x)=(1−x2)⋅∑n=0∞snx2n+1=s0+∑n=1∞(sn−sn−1)x2n+1=f(x). \begin{align*} h(x) - x^2 \cdot h(x) &= (1 - x^2) \cdot \sum_{n = 0}^\infty s_n x^{2n + 1} \\ &= s_0 + \sum_{n = 1}^\infty ( s_n - s_{n - 1} ) x^{2n + 1} = f(x). \end{align*}

Given ϵ>0\epsilon > 0, ∃N∈N\exists N \in \mathbb{N}, such that

∣sn−A∣≤ϵ,∀n≥N. |s_n - A| \le \epsilon, \quad \forall n \ge N.

Then for each x∈[0,1)x \in [0, 1),

∣(1−x2)⋅∑n=N∞snx2n+1−(1−x2)⋅∑n=N∞A⋅x2n+1∣≤ϵ⋅(1−x2)⋅11−x2≤ϵ. \begin{align*} \left| (1 - x^2) \cdot \sum_{n = N}^\infty s_n x^{2n + 1} - (1 - x^2) \cdot \sum_{n = N}^\infty A \cdot x^{2n + 1} \right| \le \epsilon \cdot (1 - x^2) \cdot \frac{1}{1 - x^2} \le \epsilon. \end{align*}

But

lim⁡x→1−(1−x2)∑n=N∞A⋅x2n+1=lim⁡x→1−A⋅x2N+1=A,lim⁡x→1−(1−x2)∑n=0Nsnx2n+1=0. \begin{aligned} \lim_{x \rightarrow 1^-} (1 - x^2) \sum_{n = N}^\infty A \cdot x^{2n + 1} &= \lim_{x \rightarrow 1^-} A \cdot x^{2N + 1} = A, \\ \lim_{x \rightarrow 1^-} (1 - x^2) \sum_{n = 0}^N s_n x^{2n + 1} &= 0. \end{aligned}

□\square