Let [ ? ] denote the greatest integer function and f ( x ) = lim n 1 n 3 k = 1 n [ k 2 3 x ] . Then 12 j = 1 f ( j ) is equal to ________. [2026]

Question

Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]

Smart Achievers · Accepted Answer

(2)

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

$\frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x}} < \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x}}$

$\lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3} \cdot 3^{x}} < \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 \cdot 3^{x} \cdot n^{3}}$

$\frac{1}{3^{x + 1}} < \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [\frac{k^{2}}{3^{x}}] \leq \frac{1}{3^{x + 1}}$

$\Rightarrow f (x) = \frac{1}{3^{x + 1}}$

$\Rightarrow 12 \sum_{j = 1}^{\infty} f (j) = 12 \sum_{j = 1}^{\infty} \frac{1}{3^{j + 1}} = 12 [\frac{1}{9} + \frac{1}{27} + \dots \infty]$

$= 12 (\frac{\frac{1}{9}}{1 - \frac{1}{3}}) = 2$

Solution

Q.
Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]

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Solution

Q. Let [⋅] denote the greatest integer function and f(x)=limn→∞1n3∑k=1n[k23x]. Then 12∑j=1∞f(j) is equal to ________. [2026]

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Q.
Let $[\cdot]$ denote the greatest integer function and $f (x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} [[\frac{k^{2}}{3^{x}}]] .$ Then $12 \sum_{j = 1}^{\infty} f (j)$ is equal to ________. [2026]