Let [·] denote the greatest integer function and f(x) = lim n→∞ (1/n³) Σ_{k=1}^n [k²/3^x]. Then 12 Σ

Q: Wait, why can we just ignore the $-1$ inside the summation when taking the limit?

That is a great observation! When we apply the Sandwich Theorem, we have a lower bound and an upper bound. The lower bound involves a term like $\frac{n}{n^3}$, which is $\frac{1}{n^2}$. As $n$ approaches infinity, $\frac{1}{n^2}$ vanishes to zero. Because both the lower and upper bounds converge to the same value, the middle term is squeezed into that same limit. We aren't ignoring the $-1$; we are proving it becomes negligible in the limit.

Q: How do I know when to use Riemann Sums vs. standard summation formulas?

Think about the structure of the limit. If you have a limit of a sum where the number of terms $n$ goes to infinity, and the expression inside can be written as a function of $\frac{k}{n}$ multiplied by $\frac{1}{n}$, that is a classic signal for a Riemann Sum. If the sum is finite or has a simple closed form, use the standard formulas. Here, the $n^3$ in the denominator is the giveaway that we are dealing with a continuous integral.

Q: Is the infinite sum formula $S = \frac{a}{1-r}$ always valid?

Not always! It is only valid if the absolute value of the common ratio $r$ is strictly less than one ($|r| < 1$). In this problem, our common ratio is $\frac{1}{3}$, which is indeed less than one, so the series converges beautifully. Always check that convergence condition before applying the formula!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating mountain. That is exactly what this problem feels like at first glance. We have a limit as $n 	o \infty$, a summation, and the dreaded greatest integer function $[ \cdot ]$ all wrapped into one.

But remember, in the world of JEE Advanced, every monster is just a collection of smaller, manageable parts. Let us break this down together.

### The Art of the Sandwich

When you see the greatest integer function $[y]$, your mind should immediately race to the **Sandwich Theorem**. It is the ultimate tool for trapping functions that refuse to be evaluated directly.

We know that for any real number $y$, the following inequality holds:
$$y - 1 < [y] \le y$$

By substituting $y = \frac{k^2}{3^x}$, we create a cage for our expression. We sum this inequality from $k=1$ to $n$:
$$\sum_{k=1}^{n} \left( \frac{k^2}{3^x} - 1 ight) < \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} ight] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$$

On the left, we have the sum of $\frac{k^2}{3^x} - 1$, and on the right, the sum of $\frac{k^2}{3^x}$. As we normalize by $n^3$, the term involving the $-1$ becomes $\frac{n}{n^3} = \frac{1}{n^2}$. As $n 	o \infty$, this term vanishes into the ether, leaving us with a clean, elegant limit.

### The Bridge to Calculus

Now, we arrive at the most beautiful part of the journey: the **Riemann Sum**. We have the limit of a sum, which is the very definition of a definite integral.

By splitting the $\frac{1}{n^3}$ into $\frac{1}{n} \cdot \frac{1}{n^2}$, we can rewrite the sum as:
$$f(x) = \frac{1}{3^x} \lim_{n 	o \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} ight)^2$$

This is the classic Riemann Sum form $\int_{0}^{1} t^2 dt$. The discrete sum has transformed into a continuous integral.

The calculation is straightforward:
$$\int_{0}^{1} t^2 dt = \left[ \frac{t^3}{3} ight]_{0}^{1} = \frac{1}{3}$$

Thus, our function simplifies to:
$$f(x) = \frac{1}{3^x} \cdot \frac{1}{3} = \frac{1}{3^{x+1}}$$

### The Final Convergence

We have tamed the beast! The function $f(x)$ is simply $\frac{1}{3^{x+1}}$. Now, we turn our attention to the final task: $12 \sum_{j=1}^{\infty} f(j)$.

Substituting our function, we get:
$$12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$$

Expanding this, we see the sequence $\frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots$. This is a perfect, infinite **Geometric Progression** with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.

Using the sum formula $S = \frac{a}{1-r}$, we get:
$$S = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{6}$$

Finally, multiplying by the $12$ from the original question, we get:
$$12 \cdot \frac{1}{6} = \mathbf{2}$$

We have conquered the mountain, and the view from the top is absolutely spectacular. Keep practicing, keep questioning, and remember that every complex problem is just a story waiting to be solved.

Animated Solution for Mathematics - Definite Integration: 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 .........

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Property of Greatest Integer Function

Constructing the Inequality

Normalizing by $n^{3}$

Applying the Sandwich Theorem

Transition to Riemann Sums

Defining the Integral

Computing the Integral

The Infinite Series $\sum f (j)$

Expanding the Series

Identifying the Geometric Progression

Calculating the Infinite Sum

Final Answer Calculation

The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

The Art of the Sandwich

The Bridge to Calculus

The Final Convergence

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