Question ID: #710
Let [ ] denote the greatest integer function and $f(x)=\lim_{n\rightarrow\infty}\frac{1}{n^{3}}\sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right]$. Then $12\sum_{j=1}^{\infty}f(j)$ is equal to:
Solution:
Using the property of the Greatest Integer Function: $x – 1 < [x] \le x$ $$ \frac{k^2}{3^x} - 1 < \left[ \frac{k^2}{3^x} \right] \le \frac{k^2}{3^x} $$
Summing from $k=1$ to $n$:
$$ \sum_{k=1}^{n} \left( \frac{k^2}{3^x} – 1 \right) < \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right] \le \sum_{k=1}^{n} \frac{k^2}{3^x} $$
$$ \frac{1}{3^x} \sum_{k=1}^{n} k^2 – n < \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right] \le \frac{1}{3^x} \sum_{k=1}^{n} k^2 $$
Divide by $n^3$ and take limit $n \to \infty$:
$$ \lim_{n \to \infty} \frac{1}{n^3} \left( \frac{n(n+1)(2n+1)}{6 \cdot 3^x} – n \right) < f(x) \le \lim_{n \to \infty} \frac{1}{n^3} \left( \frac{n(n+1)(2n+1)}{6 \cdot 3^x} \right) $$
$$ \frac{1}{3 \cdot 3^x} < f(x) \le \frac{1}{3 \cdot 3^x} $$
By Sandwich Theorem:
$$ f(x) = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}} $$
Now, we have to find $12\sum_{j=1}^{\infty}f(j)$:
$$ S = 12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} $$
$$ S = 12 \left( \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \dots \infty \right) $$
Using infinite G.P. sum formula $S_\infty = \frac{a}{1-r}$ where $a = \frac{1}{9}$ and $r = \frac{1}{3}$:
$$ S = 12 \left( \frac{1/9}{1 – 1/3} \right) $$
$$ S = 12 \left( \frac{1/9}{2/3} \right) = 12 \left( \frac{1}{6} \right) $$
$$ S = 2 $$
Ans. (2)
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