Let [·] denote the greatest integer function and $$f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right]$$. Then $$12 \sum_{j=1}^{\infty} f(j)$$ is equal to _______

For large $$n$$, the approximation $$\left[\frac{k^2}{3^x}\right]\approx\frac{k^2}{3^x}$$ holds since the floor function contributes at most 1 of error per term.

Substituting gives $$\frac{1}{n^3}\sum_{k=1}^{n}\frac{k^2}{3^x} =\frac{1}{n^3\cdot3^x}\cdot\frac{n(n+1)(2n+1)}{6} \to\frac{1}{3^x}\cdot\frac{2}{6} =\frac{1}{3\cdot3^x} =\frac{1}{3^{x+1}}\,. $$

The error from the floor function is bounded by $$\frac{1}{n^3}\sum_{k=1}^{n}\Bigl(\frac{k^2}{3^x}-\Bigl[\frac{k^2}{3^x}\Bigr]\Bigr) \le\frac{n}{n^3} =\frac{1}{n^2} \to0\,. $$

Hence $$f(x)=\frac{1}{3^{x+1}}\,. $$

$$\sum_{j=1}^{\infty}f(j) =\sum_{j=1}^{\infty}\frac{1}{3^{j+1}} =\frac{1}{3^2}\cdot\frac{1}{1-1/3} =\frac{1}{9}\cdot\frac{3}{2} =\frac{1}{6}\,. $$

 $$12\times\frac{1}{6}=2\,. $$