Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^3}}}{{{{\rm{e}}^{{n^4}}}}}$ converges.
Work Step by Step
To apply the Integral Test, we find the limits of integration by plotting the graph of the function $f\left( x \right) = \dfrac{{{x^3}}}{{{{\rm{e}}^{{x^4}}}}}$. From the figure attached, we see that $f$ is positive, decreasing, and continuous for $x \ge 1$. So, we evaluate the definite integral:
$\mathop \smallint \limits_1^\infty \dfrac{{{x^3}}}{{{{\rm{e}}^{{x^4}}}}}{\rm{d}}x$
Let $t = {x^4}$. So, $dt = 4{x^3}dx$.
$\mathop \smallint \limits_1^\infty \dfrac{{{x^3}}}{{{{\rm{e}}^{{x^4}}}}}{\rm{d}}x = \dfrac{1}{4}\mathop \smallint \limits_1^\infty \dfrac{1}{{{{\rm{e}}^t}}}{\rm{d}}t = \dfrac{1}{4}\left( { - \dfrac{1}{{{{\rm{e}}^t}}}} \right)|_1^\infty = \dfrac{1}{{4{\rm{e}}}}$
Since $\mathop \smallint \limits_1^\infty \dfrac{{{x^3}}}{{{{\rm{e}}^{{x^4}}}}}{\rm{d}}x$ converges, by the Integral Test, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^3}}}{{{{\rm{e}}^{{n^4}}}}}$ also converges.
