Asymptotics of $\sum _{k=1}^n \sum _{j=1}^n \frac{j k}{j+k}$

Question

I'm asked to find a simple asymptotical estimation of $\displaystyle \sum _{p=1}^n \sum _{q=1}^n \frac{p q}{p+q}$.

I rewrote the sum as $\displaystyle \sum _{k=2}^{2 n}\sum_{p+q=k}\frac{pq}{p+q}= \sum _{k=2}^{2 n} \sum _{p=1}^{k-1} \frac{p (k-p)}{k}$.

But it seems that $$\displaystyle \sum _{p=1}^n \sum _{q=1}^n \frac{p q}{p+q} \neq \sum _{k=2}^{2 n} \sum _{p=1}^{k-1} \frac{p (k-p)}{k}$$

What have I done wrong ?

Answer 1

We have: $$\sum_{p=1}^{n}\sum_{q=1}^{n}\frac{pq}{p+q}\\=\sum_{k=2}^{n}\frac{1}{k}\sum_{\nu=1}^{k-1}\nu(k-\nu)+\frac{1}{n+1}\sum_{\nu=1}^{n}\nu(n+1-\nu)+\sum_{k=1}^{n}\frac{1}{n+k}\sum_{\nu=k}^n\nu(n+k-\nu)\\=\sum_{k=2}^{n+1}\frac{k^2-1}{6}+\sum_{k=1}^n\frac{(n-k+1)(n(n-1)+4kn+k+k^2)}{6(n+k)}\\ =\frac{2}{3}(1-\log 2)\,n^3+o(n^3).$$

Answer 2

You forgot $p \le n$ and $q \le n$. So $p$ should go from $\max(1, k-n)$ to $\min(n, k-1)$.