Answer
Function satisfies conditions for integral test
Series diverges
Work Step by Step
$\dfrac {1}{\sqrt {1}\left( \sqrt {1}+1\right) }+\dfrac {1}{\sqrt {2}\left( \sqrt {2}+1\right) }+\dfrac {1}{\sqrt {3}\left( \sqrt {3}+1\right) }...+\dfrac {1}{\sqrt {n}\left( \sqrt {n}+1\right) }=\sum ^{\infty }_{n=1}\dfrac {1}{\sqrt {n}\left( \sqrt {n}+1\right) } $
In order integral test can be applied $f\left( x\right) =\dfrac {1}{\sqrt {x}\left( \sqrt {x}+1\right) } $ must be positive continuous and decreasing for $x>1$
We see that function is positive and continuous
$f'\left( x\right) =-\dfrac {1+\dfrac {1}{2\sqrt {x}}}{\left( x+\sqrt {x}\right) ^{2}} $
the function is decreasing
So integral test can be applied
$$ \int ^{\infty }_{1}\dfrac {1}{\sqrt {x}\left( \sqrt {x}+1\right) }=2\ln \left( \sqrt {x}+1\right) ]^{\infty }_{1}=\infty $$
The integral diverges, so the series diverges