Computing $\sum _{ k=1 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { k\left( k+n \right) }^{ 2 } } } } $ While solving a problem I encountered this:

$$\sum _{ k=1 }^{ \infty  }{ \sum _{ n=0 }^{ \infty  }{ \frac { 1 }{ { k\left( k+n \right)  }^{ 2 } }  }  } $$

For this I tried the following:
$$\sum _{ k=1 }^{ \infty  }{ \sum _{ n=0 }^{ \infty  }{ \frac { 1 }{ { k\left( k+n \right)  }^{ 2 } }  }  } =\sum _{ k=1 }^{ \infty  }{ \sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ { k\left( k+n \right)  }^{ 2 } }  }  } +\zeta \left( 3 \right) =\sum _{ k=1 }^{ \infty  }{ \sum _{ n=1 }^{ k }{ \frac { 1 }{ { k\left( n \right)  }^{ 2 } }  }  } +\zeta \left( 3 \right)  $$
From here I couldn't proceed. I'm new to multiple summations. Please help.
 A: The inner sum is the polygamma function of order $1$:
$$\psi^{(1)}(k)=\sum_{n=0}^{\infty}\frac{1}{(k+n)^2}\tag{1}$$
So the double sum can be written as
$$\sum_{k=1}^{\infty}\frac{\psi^{(1)}(k)}{k}=2\zeta(3)\approx2.404\tag{2}$$
where $\zeta(s)$ is the Riemann Zeta function. I obtained the latter result using WolframAlpha.
A: $$
\begin{align}
\sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2}
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{kn^2}\tag{1}\\
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac1{kn^2}\tag{2}\\
&=\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{3}
\end{align}
$$
Explanation:
$(1)$: substitute $n\mapsto n-k$
$(2)$: change order of summation
$(3)$: use the definition of $H_n$
We could finish by applying this answer with $q=2$, but let's do a bit more work. 
$$
\begin{align}
\sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2}
&=\zeta(3)+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k(k+n)^2}\tag{4}\\
&=\zeta(3)+\frac12\left(\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k(k+n)^2}+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n(k+n)^2}\right)\tag{5}\\
&=\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{nk(n+k)}\tag{6}\\
&=\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n^2}\left(\frac1k-\frac1{n+k}\right)\tag{7}\\
&=\zeta(3)+\frac12\sum_{n=1}^\infty\frac{H_n}{n^2}\tag{8}
\end{align}
$$
Explanation:
$(4)$: pull out the $n=0$ terms as $\zeta(3)$
$(5)$: swap $n\leftrightarrow k$ by symmetry
$(6)$: add the summands
$(7)$: partial fractions
$(8)$: $H_n=\sum\limits_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)$
Two times $(8)$ minus $(3)$ yields
$$
\sum_{k=1}^\infty\sum_{n=0}^\infty\frac1{k(k+n)^2}=2\zeta(3)\tag{9}
$$
