What is `lim_(ntooo)(1+2+3+......+n)/(1^(2)+2^(2)+3^(3)+.....+n^(2))` equal to ?

Options

A. 5
B. 2
C. 1
D. 0

Answer

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

### Step 1: Identify the formulas for the sums
The sum of the first \( n \) natural numbers is given by:
\[ S_n = 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

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

### Step 2: Substitute the formulas into the limit
Now we can substitute these formulas into our limit:
\[ \lim_{n \to \infty} \frac{S_n}{S_{n^2}} = \lim_{n \to \infty} \frac{\frac{n(n + 1)}{2}}{\frac{n(n + 1)(2n + 1)}{6}} \]

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

### Step 4: Evaluate the limit
As \( n \) approaches infinity, the term \( 2n + 1 \) also approaches infinity:
\[ \lim_{n \to \infty} \frac{3}{2n + 1} = 0 \]

### Conclusion
Thus, the limit is:
\[ \lim_{n \to \infty} \frac{1 + 2 + 3 + \ldots + n}{1^2 + 2^2 + 3^2 + \ldots + n^2} = 0 \]