If `Sigma_(r=1)^(n)t_(r)=(1)/(6)n(n+1)(n+2), AA n ge 1,` then the value of `lim_(nrarroo)Sigma_(r=1)^(n)(1)/(t_(r))` is equal to

Options

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

Answer

To solve the problem, we need to find the limit of the sum \( \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{t_r} \) given that \( \sum_{r=1}^{n} t_r = \frac{1}{6} n(n+1)(n+2) \).

### Step 1: Find the expression for \( t_n \)

We know that:
\[ S_n = \sum_{r=1}^{n} t_r = \frac{1}{6} n(n+1)(n+2) \]

To find \( t_n \), we can use the relationship:
\[ t_n = S_n - S_{n-1} \]

Now, calculate \( S_{n-1} \):
\[ S_{n-1} = \frac{1}{6} (n-1)n(n+1) \]

Now, substituting into the equation for \( t_n \):
\[ t_n = S_n - S_{n-1} = \frac{1}{6} n(n+1)(n+2) - \frac{1}{6} (n-1)n(n+1) \]

### Step 2: Simplify \( t_n \)

Let's simplify \( t_n \):
\[ t_n = \frac{1}{6} \left[ n(n+1)(n+2) - (n-1)n(n+1) \right] \]
\[ = \frac{1}{6} \left[ n(n+1) \left( (n+2) - (n-1) \right) \right] \]
\[ = \frac{1}{6} \left[ n(n+1)(3) \right] \]
\[ = \frac{1}{2} n(n+1) \]

### Step 3: Find \( \sum_{r=1}^{n} \frac{1}{t_r} \)

Now we can express \( \sum_{r=1}^{n} \frac{1}{t_r} \):
\[ \sum_{r=1}^{n} \frac{1}{t_r} = \sum_{r=1}^{n} \frac{1}{\frac{1}{2} r(r+1)} = 2 \sum_{r=1}^{n} \frac{1}{r(r+1)} \]

### Step 4: Simplify the sum \( \sum_{r=1}^{n} \frac{1}{r(r+1)} \)

Using partial fractions:
\[ \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} \]

Thus,
\[ \sum_{r=1}^{n} \frac{1}{r(r+1)} = \sum_{r=1}^{n} \left( \frac{1}{r} - \frac{1}{r+1} \right) \]

This is a telescoping series:
\[ = 1 - \frac{1}{n+1} = \frac{n}{n+1} \]

### Step 5: Substitute back into the limit

Now substituting back:
\[ \sum_{r=1}^{n} \frac{1}{t_r} = 2 \cdot \frac{n}{n+1} \]

### Step 6: Find the limit

Now we need to find:
\[ \lim_{n \to \infty} 2 \cdot \frac{n}{n+1} = \lim_{n \to \infty} 2 \cdot \frac{1}{1 + \frac{1}{n}} = 2 \cdot 1 = 2 \]

### Final Answer

Thus, the value of \( \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{t_r} \) is:
\[ \boxed{2} \]