# Proof verification: $\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$

While exploring the various applications of integral reduction formulae, I stumbled into what I believe to be a beautiful elementary proof of the equality

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$

I'd greatly appreciate it if someone took the time to review it!

To arrive at the desired result, we'll need to establish the following:

Lemma: For all $$n\geq 1$$,$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$$

We use mathematical induction to do this. The base case $$n=1$$ can be proven as follows:

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^2(x)\text{ }dx &= \int_{0}^{\frac{\pi}{4}}\left[\sec^2(x)-1\right]dx\\ &= \left[\tan(x)-x\right]_{0}^{\frac{\pi}{4}}\\ &= 1-\frac{\pi}{4}\\ &= -\left(\frac{\pi}{4}-1\right)\\ &= (-1)^1\left(\frac{\pi}{4}-\sum_{k=0}^{0}\frac{(-1)^k}{2k+1}\right)\\ &= (-1)^1\left(\frac{\pi}{4}-\sum_{k=0}^{1-1}\frac{(-1)^k}{2k+1}\right) \end{align*}

Thus, the equality holds for $$n=1$$. If it holds for some $$m\in\mathbb{N}$$, then it follows that

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2(m+1)}(x)\text{ }dx &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+2}(x)\text{ }dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\tan^2(x)\text{ }dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\left[\sec^2(x)-1\right]dx\\ &= \int_{0}^{\frac{\pi}{4}}\left[\tan^{2m}(x)\sec^2(x)-\tan^{2m}(x)\right]dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\sec^2(x)\text{ }dx-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right) \end{align*}

Here, we make the substitution $$u=\tan(x)$$. This gives

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2m}(x)\sec^2(x)\text{ }dx-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right) &= \int_{0}^{1}u^{2m}\text{ }du-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\ &= \frac{1}{2m+1}-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\ &= \frac{(-1)^m(-1)^m}{2m+1}-(-1)^m\left(\frac{\pi}{4}-\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\ &= (-1)^m\left(\frac{(-1)^m}{2m+1}-\frac{\pi}{4}+\sum_{k=0}^{m-1}\frac{(-1)^k}{2k+1}\right)\\ &= (-1)^m\left(-\frac{\pi}{4}+\sum_{k=0}^{m}\frac{(-1)^k}{2k+1}\right)\\ &= (-1)^{m+1}\left(\frac{\pi}{4}-\sum_{k=0}^{(m+1)-1}\frac{(-1)^k}{2k+1}\right) \end{align*}

This shows that the equality is true for $$n=m+1$$. Thus, mathematical induction implies that it is true for all $$n\geq 1$$.

With that settled, notice that

\begin{align*} \left|\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right| &= \left|(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)\right|\\ &= \left|\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx\right|\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx \end{align*}

because $$\tan^{2n}(x)\geq 0$$. We now obtain an upper bound for the integral $$\int_{0}^{\pi/4}\tan^{2n}(x)dx$$. Making the substitution $$x=\tan^{-1}(u)$$ gives

$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx=\int_{0}^{1}\frac{u^{2n}}{1+u^2}du$$

It's clear that $$u^{2n}/(1+u^2)< u^{2n}$$ except when $$u=0$$, so

$$\int_{0}^{1}\frac{u^{2n}}{1+u^2}du< \int_{0}^{1}u^{2n}\text{ }du=\frac{1}{2n+1}$$

Thus, we have that

$$\left|\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right|<\frac{1}{2n+1}$$

and since $$1/(2n+1)\to 0$$ as $$n\to\infty$$, it immediately follows from the squeeze theorem that

$$\lim_{n\to\infty}\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)=0$$

which is equivalent to $$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$. $$\blacksquare$$

Let me know what you think!

Edit: This is harmless, but should still be addressed: the inequality $$u^{2n}/(1+u^2) should have been $$u^{2n}/(1+u^2)\leq u^{2n}$$, since $$u^{2n}/(1+u^2)$$ and $$u^{2n}$$ are equal if and only if $$u=0$$. The inequality of integrals is still strict, though.

• @WillJagy $\sum_{n=0}^{\infty}\frac{(-1)^n(-1)^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(-1)}{2n+1}=-\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$ Commented Mar 16, 2021 at 0:39
• alright......... Commented Mar 16, 2021 at 0:43
• You should have already noticed that $\int_{0}^{\pi/4}\tan^{2n}x\,dx=\int_{0}^{1}(u^{2n}/(1+u^2))\,du$ and $\frac{1}{1+u^2}=\sum_{k=0}^{n-1}(-1)^{k-1}u^{2k}+(-1)^n\frac{u^{2n}}{1+u^2}$. Your identity is easier to derive by integrating this. In fact this is a standard approach to prove the Madhav-Gregory series for $\arctan x$. +1 for coming at it all by yourself. Commented Mar 16, 2021 at 2:52
• @ParamanandSingh I appreciate the compliment! It took me a while to convince myself that integrating the second identity would lead to the series I wanted to evaluate, but now that I do, I must say that it is one creative way to derive the sum! All that remains is to prove the second identity, which seems doable by summing a finite geometric series. Commented Mar 16, 2021 at 3:48
• Looks good to me! Commented Mar 16, 2021 at 4:11

In this question, was given for $$I_n=\int_0^{\frac \pi 4} \tan^n (x)\, dx$$ the result $$I_{n+1}-I_{n-1}=-\frac 1 n+\frac{1}{2} \left(\psi \left(\frac{n}{4}+1\right)-\psi \left(\frac{n}{4}+\frac{1}{2}\right)\right)$$ and using the same approach, it is simple to show that $$I_{n+1}+I_{n-1}=\frac 1 n$$ Adding them gives $$I_{n+1}=\frac{1}{4} \left(\psi \left(\frac{n}{4}+1\right)-\psi \left(\frac{n}{4}+\frac{1}{2}\right)\right)$$
So $$J_n=\int_0^{\frac \pi 4} \tan^{2n} (x)\, dx=\frac{1}{4} \left(\psi \left(\frac{n}{2}+\frac{3}{4}\right)-\psi \left(\frac{n}{2}+\frac{1}{4}\right)\right)$$