Let [ • denote the greatest integer function and f( x)=lim _ n -> infinity frac1 n^3 sum _ k=1^ n[frac k^23^ x]. Then 12 sum _ j=1^infinity f( j) is equal to \_\_\_\_

Let denote the greatest integer function and f x n arro... | null (Mathematics)

\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}-1\right)<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\\

\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}}<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}} \\

\begin{aligned} & \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^{3} \cdot 3^{x}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 \cdot 3^{x} \cdot n^{3}} \\ & \quad \frac{1}{3^{x+1}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \frac{1}{3^{x+1}} \\ & \Rightarrow f(x)=\frac{1}{3^{x+1}} \\ & \quad \Rightarrow 12 \sum_{j=1}^{\infty} f(j)=12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}}=12\left[\frac{1}{9}+\frac{1}{27}+--\infty\right] \\ & \quad=12 \cdot\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=2 \end{aligned}