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Question

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 __________.

Limit $f(x)$ Evaluation

The function is defined as \( f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \). We use the inequality \( y - 1 < [y] \le y \).

Applying this, we get bounds for the term inside the limit:

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

We evaluate the limits of the bounds as \( n \to \infty \). Using the formula \( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \), we find the limit:

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

The limit of the lower bound is also \( \frac{1}{3^{x+1}} \). By the Squeeze Theorem, \( f(x) = \frac{1}{3^{x+1}} \).

Summation $\sum f(j)$ Calculation

We need to compute the sum $S = \sum_{j=1}^\infty f(j)$:

$ S = \sum_{j=1}^\infty \frac{1}{3^{j+1}} $

This is an infinite geometric series with the first term \( a = \frac{1}{3^{1+1}} = \frac{1}{9} \) and the common ratio \( r = \frac{1}{3} \).

The sum of an infinite geometric series is given by \( S = \frac{a}{1-r} \) (where \( |r| < 1 \)).

$ S = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} $

Sum $12 \sum f(j)$

Finally, we calculate the required value:

$ 12 \times S = 12 \times \frac{1}{6} = 2 $

The result is 2.

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Similar Questions

  1. Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
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Important Questions from Integral Calculus

  1. Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
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  3. Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. 
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  4. Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
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