Answer
Divergent
Work Step by Step
Consider $f(x)=\dfrac{1}{x^{0.2}}$. This is positive, continuous and decreasing for $x \geq 1$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_1^\infty \dfrac{1}{x^{0.2}} dx= \lim\limits_{a \to \infty} \int_1^a \dfrac{1}{x^{0.2}} dx$
or, $ \lim\limits_{a \to \infty} [1.25x^{0.8}]_1^a=\lim\limits_{a \to \infty} [1.25a^{0.8}-1.25]=\infty$
Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{0.2}}$ is Divergent
