Find $\sum_{i=1}^\infty\left(\frac 1 {i^2}\sum_{j=1}^if(j,i)\right)$. 
For positive integers $m,n$, let $f(m,n)$ denote the number of positive integers which are both a multiple of $m$ and a factor of $n$. Find $\displaystyle \sum_{i=1}^\infty\left(\frac 1 {i^2}\sum_{j=1}^if(j,i)\right)$.  Hint: $\displaystyle\sum_{i=1}^\infty\frac 1 {i^2}=\frac{\pi^2} 6$.

This is a question from a maths contest. I have no idea to solve it. Do anyone have any idea? Thank you.
 A: Note that 
\begin{align}
f(m , n)= 
\begin{cases} 
0& \text{if} \, m \nmid n \\
d\left(\frac{n}{m}\right) & \text{if} \, m \mid n
\end{cases}
\end{align}
where $d(n)=\sum_{d \mid n}{1}$ denotes the number of divisors of $n$. Thus
\begin{align}
\sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j=1}^{i}{f(j,i)}\right)}& =\sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j \mid i}{d\left(\frac{i}{j}\right)}\right)} \\
& =\sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j \mid i}{d(j)}\right)} \\
& =\sum_{j=1}^{\infty}{d(j)\sum_{j \mid i, i \geq 1}{\frac{1}{i^2}}} \\
& =\sum_{j=1}^{\infty}{d(j)\sum_{k=1}^{\infty}{\frac{1}{(jk)^2}}} \\
& =\sum_{j=1}^{\infty}{\frac{d(j)}{j^2}\sum_{k=1}^{\infty}{\frac{1}{k^2}}} \\
& =\frac{\pi^2}{6}\sum_{j=1}^{\infty}{\frac{d(j)}{j^2}} \\
& =\frac{\pi^2}{6}\sum_{j=1}^{\infty}{\frac{1}{j^2}\sum_{d|j}{1}} \\
& =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\sum_{d \mid j, j \geq 1}{\frac{1}{j^2}}} \\
& =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\sum_{l=1}^{\infty}{\frac{1}{(dl)^2}}} \\
& =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\frac{1}{d^2}\sum_{l=1}^{\infty}{\frac{1}{l^2}}} \\
& =\frac{\pi^4}{36}\sum_{d=1}^{\infty}{\frac{1}{d^2}} \\
& =\frac{\pi^6}{216}
\end{align}
