Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n-1)^4}$ is convergent.
Work Step by Step
Let \[\sum_{n=1}^{\infty}\frac{1}{(3n-1)^4}=\sum_{n=1}^{\infty}a_n\]
\[\Rightarrow a_n=\frac{1}{(3n-1)^4}\]
$a_n$ is positive term and montonically decreasing.
So integral test is applicable.
Consider \[I=\int_{1}^{\infty}\frac{1}{(3x-1)^4}\;dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{(3x-1)^4}\;dx\;\;\;\;\;\;\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{1}{(3x-1)^4}dx\]
Substitute \[y=3x-1 \;\;\Rightarrow \;\; dy=3dx\;\Rightarrow dx=\frac{1}{3}dy\]
\[I_1=\frac{1}{3}\int\frac{1}{y^4}dy\]
\[I_1=\frac{1}{3}\int y^{-4}dy\]
\[I_1=\frac{1}{3}\left[\frac{y^{-3}}{-3}\right]\]
\[I_1=\left[\frac{-1}{9y^3}\right]\]
\[\Rightarrow I_1=\frac{-1}{9(3x-1)^3}\;\;\;\;\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[\frac{-1}{9(3x-1)^3}\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[\frac{-1}{9(3t-1)^3}+\frac{1}{72}\right]\]
\[ \Rightarrow I=0+\frac{1}{72}\]
\[ \Rightarrow I=\frac{1}{72}\]
Which is finite so $I$ is convergent.
By integral test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n-1)^4}$ is convergent.
