## vf321 3 years ago 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!

1. mukushla

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

2. mukushla

for $$|x|<1$$

3. vf321

how?

4. mukushla

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

5. vf321

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

6. mukushla

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

but this is not a complete proof

8. vf321

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

9. vf321

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

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

11. vf321

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

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

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

14. vf321

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

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

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

17. vf321

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.