Answer
The Ratio Test; diverges.
Work Step by Step
We apply the Ratio Test. Write ${a_k} = \dfrac{{{{\left[ {\pi \left( {k + 1} \right)} \right]}^k}}}{{{k^{k + 1}}}}$.
Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_{k + 1}}}}{{{a_k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{{\left[ {\pi \left( {k + 2} \right)} \right]}^{k + 1}}}}{{{{\left( {k + 1} \right)}^{k + 2}}}}\cdot\dfrac{{{k^{k + 1}}}}{{{{\left[ {\pi \left( {k + 1} \right)} \right]}^k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^{k + 1}}}}{{{{\left( {k + 1} \right)}^{k + 2}}}}\cdot\dfrac{{{{\left[ {\pi \left( {k + 2} \right)} \right]}^{k + 1}}}}{{{{\left[ {\pi \left( {k + 1} \right)} \right]}^k}}}$
$ = \pi \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^{k + 1}}}}{{{{\left( {k + 1} \right)}^{k + 2}}}}\cdot\dfrac{{{{\left( {k + 2} \right)}^{k + 1}}}}{{{{\left( {k + 1} \right)}^k}}} = \pi \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^{k + 1}}}}{{{{\left( {k + 1} \right)}^{k + 1}}}}\cdot\dfrac{{{{\left( {k + 2} \right)}^{k + 1}}}}{{{{\left( {k + 1} \right)}^{k + 1}}}}$
$ = \pi \mathop {\lim }\limits_{k \to \infty } {\left( {\dfrac{k}{{k + 1}}} \right)^{k + 1}}{\left( {\dfrac{{k + 2}}{{k + 1}}} \right)^{k + 1}} = \pi \mathop {\lim }\limits_{k \to \infty } {\left( {\dfrac{1}{{1 + \dfrac{1}{k}}}} \right)^{k + 1}}{\left( {\dfrac{{1 + \dfrac{2}{k}}}{{1 + \dfrac{1}{k}}}} \right)^{k + 1}} = \pi $
Since $\rho = \pi \gt 1$, by the Ratio Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left[ {\pi \left( {k + 1} \right)} \right]}^k}}}{{{k^{k + 1}}}}$ diverges.
