Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find 
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
 A: $$
\frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{1}{\log n}\sum_{m=1}^n\left(2\sum_{i=1}^m \frac{m+i}{m^3+i^3} - \frac{1}{m^2}\right)
$$
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
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{4\pi}{3\sqrt{3}}.
$$
A: I will be very non-rigorous here.
Letting $m = j+k$,
$\begin{align}
\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}
&\approx \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{m}{j^3+(m-j)^3}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{(j+(m-j))(j^2-j(m-j)+(m-j)^2)}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{m(j^2-jm+j^2+m^2-2jm+j^2)}\\
&= \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{1}{3j^2-3jm+m^2}\\
&= \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m^2}\sum_{j=1}^{m-1}\frac{1}{3(j/m)^2-3(j/m)+1}\\
&\approx \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m}\int_{0}^{1}
\frac{dx}{3x^2-3x+1}\\
&= \frac{1}{2}\int_{0}^{1}\frac{dx}{3x^2-3x+1}
\sum_{m=2}^{2n} \frac{1}{m}\\
&\approx \frac{\ln(2n)}{2} \int_{0}^{1}\frac{dx}{x^2-x+1/3} \\
\end{align}$.
To evaluate the integral,
$\begin{align}
I = \int_{0}^{1}\frac{dx}{x^2-x+1/3}
&=\int_{0}^{1}\frac{dx}{x^2-x+1/4-1/4+1/3}\\
&=\int_{0}^{1}\frac{dx}{(x-1/2)^2+1/12}\\
&=\int_{-1/2}^{1/2}\frac{dx}{x^2+1/12}\\
&=2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\
\end{align}
$.
Letting $x = y/\sqrt{12}= y/(2\sqrt{3})$
\begin{align}
I &= 2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\
&= 2\int_{0}^{\sqrt{3}}\frac{dy}{2\sqrt{3}((y^2)/12+1/12)}\\
&= \frac{12}{\sqrt{3}}\int_{0}^{\sqrt{3}}\frac{dy}{y^2+1}\\
&= \frac{12}{\sqrt{3}}\tan^{-1}(\sqrt{3})\\
&= \frac{12}{\sqrt{3}}(\pi/3)\\
&=\frac{4\pi}{\sqrt{3}}\\
&=\frac{4\sqrt{3}\pi}{3}\\
\end{align}
so the sum is about
$$ \frac{\ln(2n)}{2} \frac{4\sqrt{3}\pi}{3}
=\frac{4\sqrt{3}\pi\ln(2n)}{6} 
\approx \frac{2\sqrt{3}\pi\ln(n)}{3} 
$$
so the limit is
$\dfrac{2\sqrt{3}\pi}{3} $.
As usual, this is done off the top of my head,
doing math as $\LaTeX$,
but at least I got a reasonable limit.
Who knows, this might even be correct.
A: 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}} $.
