Convergence of the series $\sum \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)}$ I need to determine whether the following series converges or diverges:
$$ \sum_{n=1}^\infty \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)} $$
My first idea was to show that $a_n$ tends to zero, and so it does. 
My next plan was to rewrite the sum as
$$ a_n = \frac{1}{1+n\sqrt{n}} - \frac{1}{(n+1)\left( 1 + n\sqrt{n} \right)} $$
Where the firt one can be compared to a p-series, and the second one is smaller than the first one so it also converges. Is this correct? Is there an easier way to show that the series converge?

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 $x$.
