Proof of $$N=\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\mathrm{\frac{ab(3a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}}=\frac{1}{2}\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\frac{1}{4^{a+b+c}}$$

Conversation started Sep 14, 2020 at 19:53.

Safdar

Sep 14, 2020 19:53

How would you prove this?

$$N=\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\mathrm{\frac{ab(3a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}}=\frac{1}{2}\sum^\infty_{a=1}\sum^\infty_{b=1}\sum^\infty_{c=1}\frac{1}{4^{a+b+c}}$$

Astyx

Sep 14, 2020 20:03

Sum throughout all the permutations of a, b, c

This gives $3\sum\sum\sum {1\over 4^{a+b+c}}$

Then divide by the number of such permutations to get the result

@Safdar

Safdar

@Astyx you can do this because $a,b,c$ are independant of each other?

Astyx

You can do this because a,b and c are mute variables

For a simpler example

$$2\sum_{x=1}^k\sum_{y=1}^k{x\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} +\sum_{x=1}^k\sum_{y=1}^k{x\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} + \sum_{y=1}^k\sum_{x=1}^k{y\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x\over x+y} + \sum_{x=1}^k\sum_{y=1}^k{y\over x+y} = \sum_{x=1}^k\sum_{y=1}^k{x+y\over x+y} = k$$

Safdar

@Astyx so you are saying that $x, y$ are interchangable here.. Similarly in the above case, the same applies for $a,b,c$.. Have I understood you correctly?

Astyx

kind of

interchangeable is not the word

Safdar

@Astyx dummy variables?

Astyx

Sep 14, 2020 20:11

yep

Conversation ended Sep 14, 2020 at 20:11.

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