A infinite sum with harmonic serie Proof or disproof the folowing statement:

$$\sum_{n=1}^{+\infty}\frac{2n+1}{(n^2+n)^2}H_n=\sum_{n=1}^{+\infty}\frac{1}{n^3}$$

where $\displaystyle H_n=\sum_{k=1}^{n}\frac{1}{k}$.
 A: Method 1a.
Since 
$$
\begin{align}
\frac{2n+1}{(n^2+n)^2} H_n & = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} H_n \\
&= \frac{H_n}{n^2} - \frac{H_n}{(n+1)^2},\\
&= \frac{H_n}{n^2} - \left(\frac{H_{n+1}}{(n+1)^2}-\frac{1}{(n+1)^3}\right),
\end{align}
$$
then, summing from $n=1$ to $+\infty$, by telescoping you get 
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n = \frac{H_1}{1^2}+\sum_{n=1}^{\infty}\frac{1}{(n+1)^3}=\zeta(3).$$
Method 1b.
By using Abel's transformation, you get
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n = \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}=\zeta(3),$$
you have diverse proofs of the last equality, the Euler sum identity $\zeta(2,1)=\zeta(3)$, there : Thirty-Two Goldbach Variations.
Method 2.
You have
$$\frac{2n+1}{(n^2+n)^2} x^n= \frac{(n+1)^2 - n^2}{n^2(n+1)^2} x^n= \frac{x^n}{n^2} - \frac1x\frac{x^{n+1}}{(n+1)^2},$$ then, using the standard integral representation for the harmonic numbers and the standard expansion for the dilogarithm function:
$$
H_n=\int_0^1 \frac{1-x^n}{1-x} {\rm d} x, \qquad {\rm L}_2(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^2},
$$ we deduce that
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n =\int_0^1\frac{{\rm L}_2(x)}{x} {\rm d} x= [{\rm L}_3(x)]_0^1=\zeta(3).$$
A: A starting point could be that: $\dfrac{2n+1}{(n^2+n)^2} = \dfrac{(n+1)^2 - n^2}{n^2(n+1)^2} = \dfrac{1}{n^2} - \dfrac{1}{(n+1)^2}$
