Answer
Divergent
Work Step by Step
Let us consider $f(x)=\dfrac{1}{x^{0.2}}$
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 1$
Then $\int_1^\infty \dfrac{1}{x^{0.2}} dx= \lim\limits_{k \to \infty} \int_1^k \dfrac{1}{x^{0.2}} dx= \lim\limits_{k \to \infty} [1.25x^{0.8}]_1^k$
and $\lim\limits_{k \to \infty} [1.25k^{0.8}-1.25]=\infty$
Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{0.2}}$ is Divergent.