## ganeshie8 one year ago show that $\large \sum_{n=1}^{\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4}$

1. ganeshie8

spoiler hint : $$\huge \color{white}{🔭}$$

2. freckles

$\tan(\frac{\pi}{4})=1 \\ \tan(\sum \arctan(\frac{1}{2n^2}))=1 \\ \tan(\arctan(\frac{1}{2})+\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+\cdots) \text{ w.t.s this is }1$ but that might be harder to show not sure yet

3. freckles

$\tan(\arctan(\frac{1}{2})+s_1)=\frac{\tan(\arctan(\frac{1}{2}))+\tan(s_1)}{1-\tan(\arctan(\frac{1}{2})) \tan(s_1)}=\frac{\frac{1}{2}+\tan(s_1)}{1-\frac{1}{2} \tan(s_1)}\\ =\frac{1+2 \tan(s_1)}{2-\tan(s_1)}$

4. freckles

where $s_1=\arctan(\frac{1}{8})+\arctan(\frac{1}{18})+ \cdots$

5. freckles

yep seems like a dead end for me

6. ganeshie8

thats prety clever! i think that might actually telescope... im still trying on paper..

7. ganeshie8

$\arctan\frac{1}{2n^2} \\~\\= \arctan\frac{2}{(2n)^2} \\~\\= \arctan\frac{2}{1+(2n)^2-1}\\~\\=\arctan\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$

8. ganeshie8

$\arctan(\dfrac{x-y}{1+xy}) = \arctan(x)-\arctan(y)$ let $$x=(2n+1)$$ and $$y=2n-1$$

9. anonymous

this is nice too, classic

10. ganeshie8

im biased to telescoping xD but there is another way to work this using complex numbers which is pretty neat too the basic idea is to interpret the "multiplication of complex numbers" geometrically by "adding up the angles"

11. ganeshie8

$\arg(z_1z_2) = \arg(z_1) + \arg(z_2)$

12. Astrophysics

That sounds fascinating, have you written it down, can you just post a pic of it, then you don't have to write it out

13. Astrophysics

on here

14. ganeshie8

it is easy, just need to see it one time that the angles add up when we multiply a complex number by another complex number : |dw:1435211815854:dw|

15. ganeshie8

that opens us a door for writing infinite series as an infinite product and many other cool things xD

16. ganeshie8

Keeping in mind that $$\arctan(\frac{b}{a})$$ gives the angle of complex number $$a+bi$$, we have : $N=\prod_{k=1}^\infty (a_k+ib_k)~~\implies~~\arg N=\sum_{k=1}^\infty \arctan\frac{b_k}{a_k}$

17. freckles

I do like the telecoping one. That was pretty cute stuff.

18. anonymous

@freckles you may try this$\sum_{n=0}^{\infty} \text{Arccot}(n^2+n+1)$

19. freckles

$y=arccot(n^2+n+1) \\ \cot(y)=n^2+n+1 \\ \tan(y)=\frac{1}{n^2+n+1} \\ y=\arctan(\frac{1}{n^2+n+1})$ $\frac{1}{n^2+n+1}=\frac{4}{4n^2+4n+4}=\frac{4}{4n^2+4n+1+3} \\ =\frac{4}{(2n+1)^2+3}$ $=\frac{1}{(\frac{2n+1}{2})^2+\frac{3}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{3}{4}-1} \\ =\frac{1}{1+(\frac{2n+1}{2})^2+\frac{-1}{4}} =\frac{1}{1+(\frac{2n+1}{2})^2-\frac{1}{4}} \\ =\frac{1}{1+(\frac{2n+1}{2}-\frac{1}{2})(\frac{2n+1}{2}+\frac{1}{2})} \\$ $(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})=\frac{1}{2}+\frac{1}{2}=1 \\ \text{ so we have } \\arccot(n^2+n+1)=\arctan(\frac{(\frac{2n+1}{2}+\frac{1}{2})-(\frac{2n+1}{2}-\frac{1}{2})}{1+(\frac{2n+1}{2}+\frac{1}{2})(\frac{2n+1}{2}-\frac{1}{2})}) \\ arccot(n^2+n+1)=\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2})$

20. freckles

$\sum_{n=0}^\infty arcot(n^2+n+1)=\sum_{n=0}^{\infty}(\arctan(\frac{2n+1}{2}+\frac{1}{2})-\arctan(\frac{2n+1}{2}-\frac{1}{2}))$...

21. freckles

$\arctan(\frac{1}{2}+\frac{1}{2})-\arctan(\frac{1}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{3}{2}+\frac{1}{2})-\arctan(\frac{3}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{5}{2}+\frac{1}{2})-\arctan(\frac{5}{2}-\frac{1}{2}) \\ + \\ \arctan(\frac{7}{2}+\frac{1}{2})-\arctan(\frac{7}{2}-\frac{1}{2}) \\ + \\ +\cdots$ $=-\arctan(\frac{1}{2}-\frac{1}{2})+\lim_{k \rightarrow \infty}\arctan(\frac{2k+1}{2}+\frac{1}{2})$

22. freckles

$=-\arctan(0)+\lim_{k \rightarrow \infty}\arctan(\frac{2k+2}{2}) \\ =-(0)+\lim_{k \rightarrow \infty}\arctan(k+1) \\ =0+\frac{\pi}{2} \\ =\frac{\pi}{2}$ and I only wrote 0 a bunch of times because I thought it looked really pretty

23. anonymous

nice job @freckles :-)

24. freckles

only mimicked what @ganeshie8 did in the previous problem but thank you kindly @mukushla