Proof of Leibniz $\pi$ formula

Question

I found the following proof online for Leibniz's formula for $\pi$:

$$\frac{1}{1-y}=1+y+y^2+y^3+\ldots$$ Substitute $y=-x^2$:

$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots$$

Integrate both sides:

$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$

Now plug in $1$ for $x$:

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots$$

The thing I am confused about is that the Taylor expansion for $\frac{1}{1-y}$ only works for $-1<y<1$. Why is this still a legitimate proof? At the end, we compute the integral of both sides to $y=-1$.

Answer 1

Here is a less hand-wavey formulation of what they're saying (as far as I can tell): We have two functions $f, g:(-1, 1) \to \Bbb R$ given by $$ f(x) = \frac1{1+x^2}\\ g(x) = 1-x^2 + x^4 - x^6 + \cdots $$ an they happen to be equal. Note that for any $a \in (-1,1)$, this means that $$ \left(\vphantom{\int}\arctan(a) =\right) \int_0^a f(x)\, dx = \int_0^ag(x)\,dx\left( = a - \frac{a^3}{3} + \frac{a^5}{5} - \frac{a^7}{7} + \cdots \right) $$ Now take the limit as $a \to 1$ on both sides. The left side clearly becomes $\arctan(1)$, while the right side seems to become $1-\frac13+\frac15+\cdots$. The fact that it indeed does needs a theorem in its own right, and is addressed other answers.

Answer 2

Yes, you need an extra step. You need to know this: If $$ \arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots $$ holds for $-1 < x < 1$ and the series on the right side converges at $x=1$, then the equation also holds at $x=1$.

Arthur beat me to doing it.

Answer 3

This is a really good question. This issue is very often ignored in online resources, which makes the proofs incomplete.

The key here is Abel's theorem. It states that if the function $F(x)$ is defined by a power series $$\sum_{n=1}^\infty a_nx^n$$ on the interval $(-1,1)$ and the series $$\sum_{n=1}^\infty a_n$$ converges to a number $A$, then the limit $$\lim\limits_{x\rightarrow 1^-}F(x)$$ exists and is equal to $A$ (note that in this particular case we know the limit of $F(x)=\arctan(x)$ exists, and what we care about is the equality).

This theorem bears some similarity to the theorem which states that $F(x)$ is continuous on the interval $(-1,1)$, and indeed it says that as long as $F(1)$ is defined, the function is also continuous at $1$. However, this theorem is more subtle, since the convergence in the neighbourhood of $1$ need not be absolute nor uniform.

Answer 4

Start with \begin{equation} \int \limits_0^x \frac{dt}{1+t^{2}}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots\qquad + \frac{x^{4n+1}}{4n+1}-R_{n}(x), \quad x \in \mathbb{R}: 0\leq x \leq 1 \end{equation} Where \begin{equation} R_{n}(x) = \int \limits_0^x\frac{t^{4n+2}}{1+t^{2}}dt \end{equation} NOw, since the square of a real number is certainly non-negative then $1 \leq 1+t^{2}$ hence \begin{equation} 0 \leq R_{n}(x) \leq \int \limits_0^x t^{4n+2}dt \end{equation} Or, alternatively \begin{equation} 0 \leq R_{n}(x) \leq \frac{x^{4n+3}}{4n+3} \end{equation} As, a the outset we said $0 \leq x \leq 1$ we find $$\frac{x^{4n+3}}{4n+3} \leq \frac{1}{4n+3}$$ thus $$0 \leq R_{n}(x) \leq \frac{1}{4n+3}$$ It should be clear that $$\frac{1}{4n+3} \rightarrow 0, n \rightarrow 0$$ Which clearly implies that \begin{equation} \int \limits_0^x \frac{dt}{1+t^{2}}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots \end{equation} Which then allows you to apply your reasoning toward the end to finish with your identity for $\frac{\pi}{4}.$