2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What contest is this by the way? $\endgroup$ Mar 26, 2013 at 9:47
  • $\begingroup$ It's just a small regional contest. $\endgroup$
    – JSCB
    Mar 26, 2013 at 9:49

1 Answer 1

5
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.