Let [ cdot ] denote the greatest integer func | vedclass

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(B) Using the property $x-1 < [x] \leq x$, we have: $\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \le

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Using the property $x-1 Dividing by $n^3$ and taking the limit as $n 	o \infty$:$\lim_{n 	o \infty} \frac{1}{n^3} \sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 ight) Since $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, we have $\lim_{n 	o \infty} \frac{n(n+1)(2n+1)}{6n^3 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$.By the squeeze theorem, $f(x) = \frac{1}{3^{x+1}}$.Now, $12 \sum_{j=1}^{\infty} f(j) = 12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{9} + \frac{1}{27} + \dots ight)$.This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.Sum $= 12 \left( \frac{1/9}{1 - 1/3} ight) = 12 \left( \frac{1/9}{2/3} ight) = 12 \left( \frac{1}{6} ight) = 2$.

Let $[ \cdot ]$ 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 ........... .

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