∑ m = 1 ∞ ∑ n = 1 ∞ m 3 2 n 4 m ( m 6 n + n 8 m ) {\displaystyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\frac {m^{3}2n}{4^{m}\left(m6^{n}+n8^{m}\right)}}}
