Applying the ratio test to the series
\sum_{k=1}^\infty \frac{3^{k}}{k!},
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. Your answer should not
contain factorials, exponents, or multiplication
symbols.

Hence the series

Note that you will have to simplify your answer for

You can earn partial credit on this problem.