Question

I came across a theorem ,
but how to prove it

Answer 1

The sum of the reciprocals of the squares of the positive integers is
\[\frac{ \pi ^{2} }{ 6 }\]

Answer 2

\[\large \dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots = \dfrac{\pi^2}{6}\]

Answer 3

@ikram002p

Answer 4

Yes how to prove it

Answer 5

ermm , this need special types of series to approximate functions
the simplest way is fourier series , familour with it ?

Answer 6

nope if it is fourier then it's okay i am not familiaar

Answer 7

i'll give u the direct solution to have an idea about it :D

Answer 8

I don't want i won't understand anyways

Answer 9

\(f(x)=x^2\)
fourier series of this function is given by

\(f(x)=\dfrac{\pi^2}{3}-4 \cos x + \cos 2x -\dfrac{4}{9} cos 3x +\dfrac{4}{16} cos 4x -\dfrac{4}{25} cos 5x +.....\)

i think then it will be easy just think tricky about it from here :P

Answer 10

@ganeshie8 do you know how to get 1/n^2 sum from here ?

Answer 11

nope, il need to review

Answer 12

no need to review u have the unction already
just need to find x that gives patern of 1/n^2

Answer 13

\(f(x)=\dfrac{\pi^2}{3}+4(- \cos x + \cos 2x -\dfrac{1}{9} cos 3x +\dfrac{1}{16} cos 4x -\dfrac{1}{25} cos 5x +.....)\)

we need to find x such that cos nx =-1 when n is odd and cos nx =1 when n is even

Answer 14

can u think of one ?