# 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.

• What contest is this by the way? Mar 26, 2013 at 9:47
• It's just a small regional contest.
– JSCB
Mar 26, 2013 at 9:49

where $d(n)=\sum_{d \mid n}{1}$ denotes the number of divisors of $n$. Thus