Consider the series \(\sum _{k=1}^{\infty } (-1)^k \frac {3k^2}{2k^2+4}\). Then:
The series diverges by the divergence test. The series converges by the divergence test. The divergence test is inconclusive.

Indeed, since \(\lim _{n \to \infty } (-1)^n \frac {3n^2}{2n^2+4} = \answer {DNE}\), the series diverges by the divergence test. (Use \(+\infty \) or \(-\infty \) if appropriate or write “DNE" if the limit does not exist otherwise)

Consider the sequence \(\{b_n\}_{n=1}\) given by \(b_n = \sin \left (\frac {1}{n}\right )\). Consider the series \(\sum _{k=1}^{\infty } b_k\):
The series diverges by the divergence test. The series converges by the divergence test. The divergence test is inconclusive.

Indeed, since \(\lim _{n \to \infty } b_n = \answer {0}\), so the divergence test is inconclusive.

Consider the series \(\sum _{k=1}^{\infty } \left (\frac {2k+1}{3k+\sqrt {k}}\right )^4\). Then:
The series diverges by the divergence test. The series converges by the divergence test. The divergence test is inconclusive.

Indeed, since \(\lim _{n \to \infty } \left (\frac {2n+1}{3n+\sqrt {n}}\right )^4 = \answer {\frac {16}{81}} \neq 0\), the series diverges by the divergence test.

Note that \(\lim _{n \to \infty }\frac {2n+1}{3n+\sqrt {n}} = \answer {\frac {2}{3}}\), so \(\lim _{n \to \infty } \left (\frac {2n+1}{3n+\sqrt {n}}\right )^4 = \left (\answer {\frac {2}{3}}\right )^4 \).