Infinite summation formulas involving e, pi and integer multiples
`1 = 24 sum_(n=1)^oo (n)/(e^(\pi n) + 1) - 48 sum_(n=1)^oo (n)/(e^(2 \pi n) + 1)`
`1 = 24 sum_(n=1)^oo (n^3)/(e^(\pi n) - 1) - 264 sum_(n=1)^oo (n^3)/(e^(2 \pi n) - 1)`
`1 = 24 sum_(n=1)^oo (n^3)/(e^(\pi n) + 1) - 216 sum_(n=1)^oo (n^3)/(e^(2 \pi n) - 1)`
`1 = 132 sum_(n=1)^oo (n^3)/(e^(\pi n) + 1) - 108 sum_(n=1)^oo (n^3)/(e^(\pi n) - 1)`
`1 = 252 sum_(n=1)^oo (n^5)/(e^(\pi n) - 1) - 252 sum_(n=1)^oo (n^5)/(e^(\pi n) + 1)`
`1 = 4 sum_(n=1)^oo (n^7)/(e^(\pi n) - 1) - 484 sum_(n=1)^oo (n^7)/(e^(2 \pi n) - 1)`
`1 = 4 sum_(n=1)^oo (n^7)/(e^(\pi n) + 1) - 476 sum_(n=1)^oo (n^7)/(e^(2 \pi n) - 1)`
`1 = 242 sum_(n=1)^oo (n^7)/(e^(\pi n) + 1) - 238 sum_(n=1)^oo (n^7)/(e^(\pi n) - 1)`
`1 = 132 sum_(n=1)^oo (n^9)/(e^(\pi n) - 1) - 132 sum_(n=1)^oo (n^9)/(e^(\pi n) + 1)`
Found and proven by Dr. Stefan Gruenwald (July, 2016)