Question

show that \[\large \sum\limits_{b=1}^{\infty}\sum\limits_{a=1}^{\infty}\dfrac{1}{a^2-b^2} = \dfrac{\pi^2}{8}\] \(a\ne b\)

Answer 1

=D This looks fun!

Answer 2

Indeed it is xD my first insticts are to try partial fractions but i have no clue how to proceed really

Answer 3

Difference of squares seems like my first instinct, cause that'll look kind of like a geometric series hmmm

Answer 4

ive updated the qestion slightly..

Answer 5

\[\large \sum\limits_{b=1}^{\infty}\sum\limits_{a=1}^{\infty}\dfrac{1}{a^2-b^2} =\sum\limits_{b=1}^{\infty} \left(\dfrac{1}{2b}\sum\limits_{a=1}^{\infty}\dfrac{1}{a-b}-\dfrac{1}{a+b}\right)\]

Answer 6

idk if that telescopes ever but thats as far as i can get on my own xD

Answer 7

Summation is a linear operator so you need to use partial fractions

Answer 8

Sorry thought your reply is what you had to show, let me rethink lol

Answer 9

The \(a\) sum apparently has a closed form, \(\dfrac{\tan{b\pi}-b\pi}{b\tan{b\pi}}\).

Answer 10

Wait the question has to be wrong since a=1 and b=1 is a term.

Answer 11

it is given as a caveat that the sum doesn't include \(a = b\) terms

Answer 12

Oh sorry I see, in the question the latex sometimes breaks for the first question so I sort of missed that.

Answer 13

from the s&g formula, \(\tan(b\pi) = 0\) for all \(b \ge 1\)... denominator becomes 0 hmm

Answer 14

Hmm maybe I'm wrong, but it seems like the sum should be zero, since there will always be one matching term, like a=2 and b=1 has an a=1 and b=2 term to cancel it out. \[\Large \frac{1}{2^2-1^2}+\frac{1}{1^2-2^2}=\frac{1}{3}-\frac{1}{3} = 0\]

Answer 15

Maybe there's a way to rewrite the summation as a double integral?

Answer 16

how about looking at the continous version of it