WeBWorK Standalone Renderer

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 .

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

You can earn partial credit on this problem.