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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 $y=-1$.

Hrhm
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    The reason why it works is that when we integrate a power series, depending on the function we can "gain" the endpoints of the inteval, if it was open. On the other side it could happen that we can "lose" the endpoints when we derivate, if it was closed. Therefore we usually check the endspoints separately when we integrate.The formal prove comes from Taylor's Theorem – Stefan4024 Jun 15 '16 at 15:10
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    see this answer http://math.stackexchange.com/a/1820929/72031 This approach treats the corner case $x = \pm 1$ of the series for $\arctan x$ on the same footing as the usual case $|x| < 1$. The above approach in your post involves the integration of power series which is valid but needs to justified at least for beginners who aren't aware that it needs justification. Moreover this approach will always leave the boundary points (like $x = \pm 1$ here) of the interval/circle of convergence of the series involved. – Paramanand Singh Jun 15 '16 at 16:36

4 Answers4

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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.

Wojowu
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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 $\frac{\pi}{4}.$

Robert Z
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  • Could you explain where you got the first equation? – Hrhm Jun 15 '16 at 16:15
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    Bacon's first equation is obtained from $\sum_{j=0}^n(-y^2)^j=$ $(1-(-y^2)^{n+1})/(1+y^2)=$ $=(1+y^2)^{-1}+(-1)^n y^{2 n+2}/(1+y^2) $ by integrating this from $y=0$ to $y=x$. – DanielWainfleet Jun 15 '16 at 19:01
  • @Hrhm user254665 is correct, this is where I take the first step, I hope that is useful for you. –  Jun 16 '16 at 07:51
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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.

Arthur
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    There is still a nontrivial point left, namely that $a-a^3/3+a^5/5-\dots\rightarrow 1-1/3+1/5-\dots$. I think you might want to mention that in your answer. – Wojowu Jun 15 '16 at 16:05
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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.

GEdgar
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