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