A not so familiar summation

The summation could be expressed as

$A = \large \lim\limits_{n \to \infty} \sum\limits_{k=1}^{n} \frac{ \ln (1+ \frac{k}{n})}{k}$

We could modify a little bit of the RHS and get

$A = \large \lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{n} \frac{ \ln (1+ \frac{k}{n})}{\frac{k}{n}}$

Well, this is actually the Riemann Sum representation of the integral

$A = \large \int_0^1 \frac{ \ln(1+x) }{x} \mathrm{d}x$

Which, at first glance, isn't computable.

Let us turn to power series for help.

Note that $\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}...$

$\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}...$

$\frac{\ln (1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \frac{x^4}{5}...$

$\int_0^x \frac{\ln (1+x)}{x} = x - \frac{x^2}{4} + \frac{x^3}{9} - \frac{x^4}{16} + \frac{x^5}{25}...$

Substituting $x=1$ to the above equation, we get an infinite sum equating to $A$ .

Now, we know of this sum

$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +...$

$\frac{\pi^2}{12} = \frac{1}{2} + \frac{1}{8} + \frac{1}{18} + \frac{1}{32} +...$

$\frac{\pi^2}{6} - \frac{\pi^2}{12} = 1 + (\frac{1}{4} - \frac{1}{2}) + \frac{1}{9} + (\frac{1}{16} - \frac{1}{8}) + \frac{1}{25} + ( \frac{1}{36} - \frac{1}{18} ) + ...$

$\frac{\pi^2}{12} = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36} +... = A$

So that's it! $A = \frac{\pi^2}{12}$ , such that

$\lfloor 10^4 A \rfloor = \boxed{8224}$