Answer
$\displaystyle\sum_{n=1}^{\infty}n^{-0.3}$ is divergent
Work Step by Step
Let \[\sum_{n=1}^{\infty}n^{-0.3}=\sum_{n=1}^{\infty}a_n\]
\[\Rightarrow a_n=n^{-0.3}=\frac{1}{n^{0.3}}\]
$a_n$ is positive term and montonically decreasing.
So integral test is applicable.
Consider \[I=\int_{1}^{\infty}x^{-0.3}\;dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-0.3}\;dx\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[\frac{x^{0.7}}{0.7}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow\infty}\left[\frac{t^{0.7}}{0.7}-\frac{1}{0.7}\right]\]
\[I=\infty\]
$\Rightarrow I$ is divergent.
By integral test $\displaystyle\sum_{n=1}^{\infty}n^{-0.3}$ is divergent.
