one can apply the Ratio Test:
lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( n + 1 ) 4 2 ( n + 1 ) / 3 / ( 2 ( n + 1 ) + 7 ) 4 n 4 2 n / 3 / ( 2 n + 7 ) 4 | = lim n → ∞ ( n + 1 ) 4 ( 2 n + 7 ) 4 n 4 ( 2 n + 9 ) 4 2 ( n + 1 ) / 3 2 n / 3 = 1 ⋅ 2 1 / 3 > 1. {\displaystyle {\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {(n+1)^{4}2^{(n+1)/3}/(2(n+1)+7)^{4}}{n^{4}2^{n/3}/(2n+7)^{4}}}\right|\\&=\lim _{n\to \infty }{\frac {(n+1)^{4}(2n+7)^{4}}{n^{4}(2n+9)^{4}}}{\frac {2^{(n+1)/3}}{2^{n/3}}}=1\cdot 2^{1/3}>1.\end{aligned}}}
Therefore, the series diverges.
