## vf321 Group Title Why does 1-1/3+1/5-1/7... converge to Pi/4? In Latex, $\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$ Don't give me a copy/paste of what Wikipedia says though, it skips a couple steps and I don't understand! 2 years ago 2 years ago

1. mukushla Group Title

sure u know that$\arctan x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$

2. mukushla Group Title

for $$|x|<1$$

3. vf321 Group Title

how?

4. mukushla Group Title

$\frac{1}{1+t^2}=1-t^2+t^4-t^6+t^8-...=\sum_{n=0}^{\infty} (-1)^n t^{2n} \ \ \ \ \ \ \ |t|<1$integrate both sides from $$0$$ to $$x$$ for $$|x|<1$$

5. vf321 Group Title

Wow should have caught that more easily. It seems so clear when you see it. But go on with the proof....

6. mukushla Group Title

after integrating the geometric series u will get$\arctan x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$let $$x=1$$ u will have$\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^n }{2n+1}$

7. mukushla Group Title

but this is not a complete proof

8. vf321 Group Title

Yeah but you can't just let x=1 cause that series only holds true for $|x|<1$

9. vf321 Group Title

So really the question is how Wikipedia got that first substitution of arctanx with the series that had a larger radius of convergence. I get the rest.

10. mukushla Group Title

here we must use Abel's theorem that im not comfortable with

11. vf321 Group Title

Well can we try to reason this out? Clearly the introduction of $\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}$in the summation $\frac{1}{1+x^2}=\sum_{k=0}^n(-1)^nx^{2k}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}$Had something to do with it...

12. vf321 Group Title

My guess is that Wikipedia made that sum finite (with n on top instead of infy), and so that messy term at the end is just a correction factor for when x=1 in the mclauren series 1/(1+x^2). Then, taking the limit n->infty later guarentees that the new series models arctan x, and, after integration, the correction factor approaches 0 at infity.

13. vf321 Group Title

So integration actually expanded our radius of convergence (as I understand it)

14. vf321 Group Title

If I'm not mistaken, then it looks something like this: $\frac{1}{1+x^2}=\sum_{k=0}^n (-1)^kx^{2k}+R$where$R=\{_{0,|x|<1}^{f(x, k), x=1}$We know that there must be some existing f(x, k) for finite series. We also know, because of the radius of convergencIe for a geometric series, that $\lim_{n\rightarrow\infty}\sum_{k=0}^nf(x, k)=0$If and only if |x|<1, and for x=1, the same sum diverges.

15. vf321 Group Title

However, integrating both sides of my first equation in the previous reply gets us: $arctan(x)=\sum_{k=0}^n\int(-1)^kx^{2k}dx+\int Rdx$ Taking the limit n->infty of both sides, we find that, due to the properties of f(x, k), $\lim_{n\rightarrow\infty}\int R dx=0$ For -1 < x <= 1. Then your proof works.

16. vf321 Group Title

EDIT: Last limit should be: $\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0$

17. vf321 Group Title

The whole thing is interesting, though. It leads me to another question, that I'll post separately (about whether or not such a function f(x, k) can exist with those properties.