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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 .
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.

You can earn partial credit on this problem.