<s>[INST] Limit of double sum: $\lim\limits_{n\to\infty}n^{-2}\sum\limits_{k=1}^n\sum\limits_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$ Who is so kind to enlighten me about the steps I need to follow? 
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\sum_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$$
 [/INST] We have a Riemann sum:
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \left ( \frac{1-\frac{2k}{n}}{1+\frac{2k}{n}}\right)^2 \sum_{m=k+1}^n \frac{1-\frac{2m}{n}}{1+\frac{2m}{n}} = \int_0^1 dx \left (\frac{1-2 x}{1+2 x}\right )^2 \, \int_x^1 dy \frac{1-2 y}{1+2 y}$$
The evaluation of the above integral is straightforward but messy.  The inner integral has an antiderivative
$$\begin{align}\int dy \frac{1-2 y}{1+2 y} &= \int \frac{dy}{1+2 y} - \int dy \frac{2 y}{1+2 y}\\ &= \frac12 \log{(1+2 y)} - \left [y - \frac12 \log{(1+2 y)} \right ]\\ &= \log{(1+2 y)}-y\end{align}$$
The integral is now a single integral when the inner integral is evaluated over its integration limits:
$$\int_0^1 dx \left (\frac{1-2 x}{1+2 x}\right )^2 \left [\log{3} - 1 + x - \log{(1+2 x)} \right ]$$
This integral may be evaluated by substituting $u=1+2 x$, $x=(u-1)/2$, to get
$$\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \left [\log{3}-1+\frac{u-1}{2} - \log{u}\right]$$
Now,
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right )(\log{3}-1) &= \left(\frac{7}{3}-2 \log{3}\right) ( \log{3}-1) \\ &= -\frac{7}{3} + \frac{13}{3} \log{3} - \log^2{3}\end{align}$$
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \log{u} &= 2 \int_1^3 du \frac{\log{u}}{u^2} - 2 \int_1^3 du \frac{\log{u}}{u} + \frac12 \int_1^3 du \, \log{u} \\ &= 2 \left [ - \frac{\log{u}}{u}\right]_1^3 + 2 \int_1^3 \frac{du}{u^2} - [\log^2{u}]_1^3 + \frac12 [u \log{u}-u]_1^3 \\ &= \frac13 + \frac{5}{6} \log{3} - \log^2{3}\end{align}$$
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \frac{u-1}{2} &= \frac14 \int_1^3 du \left (-\frac{4}{u^2} + \frac{8}{u} - 5+u \right )\\ &= 2 \log{3}-\frac{13}{6}\end{align}$$
Adding the above three results together, I get for the desired limit:
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \left (\frac{n-2 k}{n+2 k} \right )^2 \, \sum_{m=k+1}^n \frac{n-2 m}{n+2 m} = -\log^2{3} + \frac{33}{6} \log{3} - \frac{29}{6}$$
which is about $0.002085$.
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