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

$$
\frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{1}{\log n}\sum_{m=1}^n\left(2\sum_{i=1}^m \frac{m+i}{m^3+i^3} - \frac{1}{m^2}\right)
$$
Since
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m^2} = 0,
$$
we conclude
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = 2 \lim_{n\to\infty}\frac{1}{\log n}\sum_{m=1}^n\left(\sum_{i=1}^m \frac{m+i}{m^3+i^3}\right).
$$
Define
$$
a_m = m \cdot \sum_{i=1}^m \frac{m+i}{m^3+i^3},
$$
so 
$$
(1)\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \lim_{n\to\infty} 2 \frac{1}{\log n} \sum_{m=1}^n \frac{a_m}{m}.
$$
We can approximate $a_m$ using a Riemann integral, giving
$$
(2) \lim_{m\to\infty} a_m = \int_0^1 \frac{1+x}{1+x^3} = \frac{2\pi}{3\sqrt{3}}.
$$
(I forget how to do that last integral. I used Mathematica.) Recall that
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m} = 1.
$$
For any sequence $b_m\to 0$,
$$
(3) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = 0.
$$
To see this, suppose $|b_m|<\epsilon$ for $m>N$. Then
$$
\left|\frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m}\right| <= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \left|\frac{1}{\log n}\sum_{m>N}^n b_m \frac{1}{m}\right| \\
   <= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \frac{\epsilon}{\log n}\sum_{m>N}^n \frac{1}{m}.
$$
The limit of the right hand side of this inequality is $\epsilon$. Since $\epsilon$ can be chosen arbitrarily small, (3) follows. As an immediate corollary, if $b_m\to L$, then
$$
(4) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = L.
$$
It follows from (1), (2), and (4) that
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{4\pi}{3\sqrt{3}}.
$$
