Answer
The Ratio Test; converges.
Work Step by Step
We apply the Ratio Test. Write ${a_k} = {k^{50}}{{\rm{e}}^{ - k}} = \dfrac{{{k^{50}}}}{{{{\rm{e}}^k}}}$.
Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_{k + 1}}}}{{{a_k}}} = \dfrac{{{{\left( {k + 1} \right)}^{50}}}}{{{{\rm{e}}^{k + 1}}}}\cdot\dfrac{{{{\rm{e}}^k}}}{{{k^{50}}}} = \dfrac{1}{{\rm{e}}}\mathop {\lim }\limits_{k \to \infty } {\left( {\dfrac{{k + 1}}{k}} \right)^{50}} = \dfrac{1}{{\rm{e}}}\mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^{50}} = \dfrac{1}{{\rm{e}}}$
Since $\rho = \dfrac{1}{{\rm{e}}} \lt 1$, by the Ratio Test (Theorem 9.5.5), the series $\mathop \sum \limits_{k = 1}^\infty {k^{50}}{{\rm{e}}^{ - k}}$ converges.
