For deciding convergence, Siminore's comparison test answers affirmatively. If for real $x>0$ we wish to approximate the sum over $\mathbb{Z}^+$ of $ \eqalign{ f(x) &=\frac{x}{(x+1)(x^{3/2}+1)} =\left(1-\frac1{1+x}\right)\left(1+x^{3/2}\right)^{-1}\\ &=\frac1{1+x^{3/2}} -\frac1{\left(1+x\right)\left(1+x^{3/2}\right)} =\frac1{\left(1+x^{-1}\right)\left(1+x^{3/2}\right)} } $ then its antiderivative is $ F(x) = \arctan\left(\sqrt{x}\right) + \frac16 \, \log \frac{ \left(x - \sqrt{x} + 1\right)^4 }{ \left(x + 1\right)^3 \left(\sqrt{x} + 1\right)^2 } $ and a rough lower bound on the sum woud be $ S = \sum_{x=1}^\infty f(x) > I = \int_1^\infty f(x)\,dx = \frac{\pi}4 + \frac56 \, \log2 \approx 1.363021 $ since $f(x)$ is strictly decreasing for $x > 0.73173541$

since its global maximum is at the root $t\approx0.85541534$ of the denominator $3t^5+t^3-2$ (for $t=\sqrt x$) of $ f'(x)= -\frac{ 3x^{5/2}+x^{3/2}-2 }{ 2\,\left(x+1\right)^2 \left(x^3+2x^{3/2}+1\right)} \,. $ We can think of our sum $S$ as a left-endpoint (improper) Riemann sum. A correspondingly rough upper bound would be $f(1)+I=\frac14+I\approx1.613021$. A better approximation would be the average of these two: $ S \approx \frac18 + \frac{\pi}4 + \frac56 \, \log2 \approx 1.488021 \,. $ The antiderivative $F$ above can be derived by setting $x=t^2$ (so $dx=2t\,dt$) and $ g(t)=f(t^2)=\frac{t^2}{(t^2+1)(t^3+1)} $ so that $ \eqalign{ t\,g(t)& =\tfrac13\frac{2t-1}{t^2-t+1} -\tfrac12\frac{t-1}{t^2+1} -\tfrac16\frac1{t+1} \\& =\tfrac13\frac{2t-1}{t^2-t+1} -\tfrac14\frac{2t}{t^2+1} -\tfrac16\frac1{t+1} +\tfrac14\frac1{t^2+1} } $ and, for $t>0$ (to dispense with absolute values inside the natural logarithm), $ \eqalign{ F(x)& = \int f(x)\,dx =2\int t\,g(t)\,dt \\& = \frac23\ln\left(t^2-t+1\right) - \frac12\ln\left(t^2 +1\right) - \frac13\ln\left(t +1\right) + \frac12\arctan{t} } $ which yields the above in $x$.