Question

show that \[\large \sum_{n=1}^{\infty}\arctan\frac{1}{2n^2}=\frac{\pi}{4}\]

Answer 1

spoiler hint :
\(\huge \color{white}{🔭}\)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yep seems like a dead end for me

Answer 6

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

Answer 7

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

Answer 8

\[\arctan(\dfrac{x-y}{1+xy}) = \arctan(x)-\arctan(y)\]

let \(x=(2n+1)\) and \(y=2n-1\)

Answer 9

this is nice too, classic

Answer 10

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"

Answer 11

\[\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\]

Answer 12

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

Answer 13

on here

Answer 14

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 :