We apply the Limit Comparison Test with
a n = n 4 2 n / 3 ( 2 n + 7 ) 4 and b n = 2 n / 3 , {\displaystyle a_{n}={\frac {n^{4}2^{n/3}}{(2n+7)^{4}}}\quad {\text{and}}\quad b_{n}=2^{n/3},}
which is valid since lim n → ∞ a n b n = lim n → ∞ n 4 ( 2 n + 7 ) 4 = lim n → ∞ 1 ( 2 + 7 / n ) 4 = 1 2 4 {\displaystyle {\begin{aligned}\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\rightarrow \infty }{\frac {n^{4}}{(2n+7)^{4}}}=\lim _{n\rightarrow \infty }{\frac {1}{{(2+7/n)}^{4}}}={\frac {1}{2^{4}}}\end{aligned}}}
exists and is nonzero. Since the series ∑ n = 1 ∞ b n {\displaystyle \sum _{n=1}^{\infty }b_{n}} is a divergent geometric series (with common ratio r = 2 1 / 3 > 1 {\displaystyle r=2^{1/3}>1} ), the given series diverges.
The answer is diverge.
is a divergent geometric series (with common ratio 
), the given series diverges.
