Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find 
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$

Here is another sketch of proof.
Let
$$J_n = \{(j, k) : 0 \leq j, k < n \text{ and } (j, k) \neq (0, 0) \}.$$
Then for each $(j, k) \in J_n$ and $(x_0, y_0) = (j/n, k/n)$, we have
$$ \frac{x_0 + y_0}{(x_0+\frac{1}{n})^{3} + (y_0 + \frac{1}{n})^3} \leq \frac{x+y}{x^3 + y^3} \leq \frac{x_0 + y_0 + \frac{2}{n}}{x_0^3 + y_0^3} $$
for $(x, y) \in [x_0, y_0] \times [x_0 + 1/n, y_0 + 1/n]$. Thus if we let $D_n$ be the closure of the set $[0, 1]^2 - [0, 1/n]^2$, then
$$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} \leq \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy \leq \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3}. $$
It is not hard to establish the relation that
$$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1) $$
and that
$$ \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1). $$
By noting that
\begin{align*}
\int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy
&= 2\int_{\frac{1}{n}}^{1} \int_{0}^{y} \frac{x+y}{x^3 + y^3} \, dxdy \\
&= (2 \log n) \int_{0}^{1} \frac{1}{x^2 - x + 1} \, dx \\
&= \frac{4 \pi \log n}{3\sqrt{3}},
\end{align*}
we obtain the asymptotic formula
$$ \frac{1}{\log n} \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} = \frac{4 \pi}{3\sqrt{3}} + O\left( \frac{1}{\log n} \right) $$
and hence the answer is $\displaystyle \frac{4 \pi}{3\sqrt{3}} $.
