Answer
Converges
Work Step by Step
Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$
1. For $l \lt 1$, the series is absolutely convergent.
2. For $l \gt 1$, the series is divergent.
3. For $l = 1$, the series is inconclusive.
Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty} \dfrac{\alpha^{(k+1)}}{(k+1)^{\alpha}} \times \dfrac{k^{\alpha}}{\alpha^{k}} \\=\lim\limits_{k \to \infty} \dfrac{\alpha \cdot \alpha^{(k)}}{\alpha^k} \times (\dfrac{k}{k+1})^{\alpha}\\=\alpha (1-0)^{\alpha}\\=\alpha $
So, we can conclude that the given series converges when $0 \lt \alpha \lt 1$ by the ratio test.