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Find the infinite sum of f(a, b, c) = ((2 / 5) ^ a) * ((11) ^ (-c)) * ((7 ^ (-a - b))

for ordered triples(a, b, c) such that a, b, c satisfy the triangle inequality. I simplified the input to a simpler formula given above but unable to approach further. Answer can be represented in the form p / q. Here a, b, c must be positive integers.

I am unable to put this condition to calculate this sum.

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    $\begingroup$ Is this about Wolfram Mathematica or a math question? $\endgroup$ – Kuba Jan 26 '18 at 9:49
  • $\begingroup$ I tried to simplify using Wolfram Mathematica but unable to code it as I don't know how to put in the triangle inequality. Ultimately I needed the answer, so I decided to code it out as I was unable to solve it analytically. $\endgroup$ – Kunwar Singh Jan 26 '18 at 9:51
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You can use Boole:

Sum[
    ((2/5)^a)*((11)^(-c))*(7^(-a-b)) Boole[a<b<c && c<a+b],
    {a,Infinity},
    {b,Infinity},
    {c,Infinity}
]

1/1516743381

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 Sum[((2/5)^a)*(11^(-c))*(7^(-a - b)) ,
  {a, Infinity}, {b, a + 1, Infinity}, {c, b + 1, a + b - 1}]

1/1516743381

note this works for symbolic contstants..

Sum[p^a/q^c/ r^(a + b)  ,
 {a, Infinity}, {b, a + 1, Infinity}, {c, b + 1, a + b - 1}]

p^2/((-1 + q r) (-p + q r^2) (-p + q^2 r^2))

% /. {p -> 2/5, q -> 11, r -> 7}

1/1516743381

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