Applying the ratio test to the series
\sum_{k=1}^\infty \frac{3^{k}}{k^{3}},
you would compute

\displaystyle \lim_{k \to \infty} \,
\biggl \lvert \dfrac {a_{k + 1}} {a_k}
\biggr \rvert = \lim_{k \to \infty}
= .

Hence the series
converges
diverges
inconclusive
.

Note that you will have to simplify your answer for\frac{a_{k+1}}{a_k} otherwise you will
get an error message.

Hence the series

Note that you will have to simplify your answer for

You can earn partial credit on this problem.