<s>[INST] 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?
 [/INST] 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 $x$.
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