A infinite sum with harmonic serie

Question

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 $\displaystyle H_n=\sum_{k=1}^{n}\frac{1}{k}$.

Answer 1

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

$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n =\int_0^1\frac{{\rm L}_2(x)}{x} {\rm d} x= [{\rm L}_3(x)]_0^1=\zeta(3).$$

Answer 2

A starting point could be that: $\dfrac{2n+1}{(n^2+n)^2} = \dfrac{(n+1)^2 - n^2}{n^2(n+1)^2} = \dfrac{1}{n^2} - \dfrac{1}{(n+1)^2}$