`lim_(n tooo)1/n^2` (1+2+…+n) is equal to

Options

A. 0
B. 1
C. `1/2`
D. `1/4`

Answer

To solve the limit \( \lim_{n \to \infty} \frac{1}{n^2} (1 + 2 + \ldots + n) \), we can follow these steps:

### Step 1: Identify the sum of the first n natural numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ S_n = 1 + 2 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 2: Substitute the sum into the limit expression
Now, we substitute \( S_n \) into the limit:
\[ \lim_{n \to \infty} \frac{1}{n^2} \left( \frac{n(n + 1)}{2} \right) \]

### Step 3: Simplify the expression
We can simplify the expression inside the limit:
\[ = \lim_{n \to \infty} \frac{n(n + 1)}{2n^2} \]
This simplifies to:
\[ = \lim_{n \to \infty} \frac{n + 1}{2n} \]

### Step 4: Further simplify the limit
Now, we can simplify \( \frac{n + 1}{2n} \):
\[ = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} \]

### Step 5: Evaluate the limit
As \( n \to \infty \), \( \frac{1}{n} \) approaches 0. Therefore:
\[ = \frac{1 + 0}{2} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to \infty} \frac{1}{n^2} (1 + 2 + \ldots + n) = \frac{1}{2} \]

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