Note that lim n → ∞ n 4 + 3 2 n 4 + n = lim n → ∞ ( n 4 + 3 ) / n 4 ( 2 n 4 + n ) / n 4 = lim n → ∞ 1 + 3 / n 4 2 + n / n 4 = 1 + lim n → ∞ 3 / n 4 2 + lim n → ∞ 1 / n 3 = 1 2 ≠ 0. {\displaystyle \lim _{n\to \infty }{\frac {n^{4}+3}{2n^{4}+n}}=\lim _{n\to \infty }{\frac {(n^{4}+3)/n^{4}}{(2n^{4}+n)/n^{4}}}=\lim _{n\to \infty }{\frac {1+3/{n^{4}}}{2+n/{n^{4}}}}={\frac {1+\lim _{n\to \infty }3/{n^{4}}}{2+\lim _{n\to \infty }1/{n^{3}}}}={\frac {1}{2}}\neq 0.}
Therefore, by the divergence test, the given series diverges.
