`lim_(n to oo) (1^(2)+2^(2)+3^(2)+…+n^(2))/(n^(3))` is equal to

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(1)/(3)`
D. `(1)/(6)`

Explanation

To solve the limit \( \lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3} \), we will follow these steps:

### Step 1: Identify the sum of squares formula
The sum of the squares of the first \( n \) natural numbers is given by the formula:
\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 2: Substitute the formula into the limit
We can substitute this formula into our limit:
\[ \lim_{n \to \infty} \frac{\frac{n(n + 1)(2n + 1)}{6}}{n^3} \]

### Step 3: Simplify the expression
Now, simplify the expression:
\[ = \lim_{n \to \infty} \frac{n(n + 1)(2n + 1)}{6n^3} \]
\[ = \lim_{n \to \infty} \frac{(n + 1)(2n + 1)}{6n^2} \]

### Step 4: Divide numerator and denominator by \( n^2 \)
Next, we divide both the numerator and the denominator by \( n^2 \):
\[ = \lim_{n \to \infty} \frac{\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)}{6} \]

### Step 5: Evaluate the limit as \( n \) approaches infinity
As \( n \) approaches infinity, \( \frac{1}{n} \) approaches 0. Thus, we have:
\[ = \frac{(1 + 0)(2 + 0)}{6} = \frac{1 \cdot 2}{6} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3} = \frac{1}{3} \]

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