`lim_(nrarroo)([r]+[2r]+ . . . + [nr])/n^2`

Answer

To solve the limit \( \lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2} \), we will break it down step by step.

### Step 1: Understanding the Greatest Integer Function
The greatest integer function, denoted as \([x]\), represents the largest integer less than or equal to \(x\). We can express \([kr]\) as:
\[ [kr] = kr - \{kr\} \]
where \(\{kr\}\) is the fractional part of \(kr\).

### Step 2: Rewrite the Sum
We can rewrite the sum in the limit:
\[ \sum_{k=1}^{n} [kr] = \sum_{k=1}^{n} (kr - \{kr\}) = \sum_{k=1}^{n} kr - \sum_{k=1}^{n} \{kr\} \]
This gives us:
\[ \sum_{k=1}^{n} [kr] = r \sum_{k=1}^{n} k - \sum_{k=1}^{n} \{kr\} \]

### Step 3: Calculate the First Sum
The sum of the first \(n\) natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
Thus, we have:
\[ \sum_{k=1}^{n} [kr] = r \cdot \frac{n(n+1)}{2} - \sum_{k=1}^{n} \{kr\} \]

### Step 4: Analyze the Second Sum
The fractional part \(\{kr\}\) is bounded between 0 and 1. Therefore, the sum \(\sum_{k=1}^{n} \{kr\}\) is at most \(n\):
\[ 0 \leq \sum_{k=1}^{n} \{kr\} \leq n \]

### Step 5: Substitute Back into the Limit
Now substituting back into our limit:
\[ \lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2} = \lim_{n \to \infty} \frac{r \cdot \frac{n(n+1)}{2} - \sum_{k=1}^{n} \{kr\}}{n^2} \]
This simplifies to:
\[ \lim_{n \to \infty} \left( \frac{r(n^2 + n)}{2n^2} - \frac{\sum_{k=1}^{n} \{kr\}}{n^2} \right) \]
Breaking this down further:
\[ = \lim_{n \to \infty} \left( \frac{r}{2} \left(1 + \frac{1}{n}\right) - \frac{\sum_{k=1}^{n} \{kr\}}{n^2} \right) \]

### Step 6: Evaluate the Limit
As \(n\) approaches infinity, \(\frac{1}{n}\) approaches 0, and \(\frac{\sum_{k=1}^{n} \{kr\}}{n^2}\) approaches 0 since \(\sum_{k=1}^{n} \{kr\}\) is bounded by \(n\):
\[ \lim_{n \to \infty} \frac{\sum_{k=1}^{n} \{kr\}}{n^2} = 0 \]
Thus, we have:
\[ \lim_{n \to \infty} \left( \frac{r}{2} \cdot 1 - 0 \right) = \frac{r}{2} \]

### Final Answer
The final result is:
\[ \lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2} = \frac{r}{2} \]