Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{n^4+4}$ is divergent
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{n^3}{n^4+4}\]
\[\Rightarrow a_n=\frac{n^3}{n^4+4}=\frac{1}{n+\frac{4}{n^3}}\]
$a_n$ is positive term and monotonically decreasing.
So Integral test is applicable.
Let \[I=\int_{1}^{\infty}\frac{x^3}{x^4+4}dx\]
\[I=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{x^3}{x^4+4}dx\;\;\;\;\;\;\;\;\;\;\ldots(1)\]
Let \[I_1=\int\frac{x^3}{x^4+4}dx\]
Substitute $y=x^4+4\;\Rightarrow\; dy=4x^3dx$
\[\frac{1}{4}dy=x^3dx\]
\[\Rightarrow I_1=\frac{1}{4}\int\frac{dy}{y}\]
\[\Rightarrow I_1=\frac{1}{4}\ln |y|\]
\[\Rightarrow I_1=\frac{1}{4}\ln (x^4+4)\;\;\;\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow\infty}\left[\frac{1}{4}\ln (x^4+4)\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow\infty}\left[\frac{1}{4}\ln (t^4+4)-\frac{1}{2}\ln 5\right]\]
\[\Rightarrow I=\infty\]
$\Rightarrow I$ is divergent.
By integral test, $\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{n^4+4}$ is divergent.
