If `sum_(r=1)^(n) a_(r)=(1)/(6)n(n+1)(n+2)` for all `nge1`, then `lim_(ntooo) sum_(r=1)^(n) (1)/(a_(r))`, is

Options

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

Answer

To solve the problem, we need to find the limit as \( n \) approaches infinity of the summation \( \sum_{r=1}^{n} \frac{1}{a_r} \) given that \( \sum_{r=1}^{n} a_r = \frac{1}{6} n(n+1)(n+2) \).

### Step 1: Find the nth term \( a_n \)
We know that:
\[ \sum_{r=1}^{n} a_r = \frac{1}{6} n(n+1)(n+2) \]
To find \( a_n \), we can express it as:
\[ a_n = \sum_{r=1}^{n} a_r - \sum_{r=1}^{n-1} a_r \]
This means we need to compute \( \sum_{r=1}^{n-1} a_r \):
\[ \sum_{r=1}^{n-1} a_r = \frac{1}{6} (n-1)n(n+1) \]
Now, substituting these into the equation for \( a_n \):
\[ a_n = \frac{1}{6} n(n+1)(n+2) - \frac{1}{6} (n-1)n(n+1) \]

### Step 2: Simplify \( a_n \)
Now, we will simplify \( a_n \):
\[ a_n = \frac{1}{6} \left[ n(n+1)(n+2) - (n-1)n(n+1) \right] \]
Factoring out \( n(n+1) \):
\[ = \frac{1}{6} n(n+1) \left[ (n+2) - (n-1) \right] \]
\[ = \frac{1}{6} n(n+1) \left[ 3 \right] \]
Thus,
\[ a_n = \frac{1}{2} n(n+1) \]

### Step 3: Find \( \sum_{r=1}^{n} \frac{1}{a_r} \)
Now we need to compute:
\[ \sum_{r=1}^{n} \frac{1}{a_r} = \sum_{r=1}^{n} \frac{2}{r(r+1)} \]
This can be simplified using partial fractions:
\[ \frac{2}{r(r+1)} = 2 \left( \frac{1}{r} - \frac{1}{r+1} \right) \]

### Step 4: Evaluate the summation
Thus,
\[ \sum_{r=1}^{n} \frac{1}{a_r} = 2 \sum_{r=1}^{n} \left( \frac{1}{r} - \frac{1}{r+1} \right) \]
This is a telescoping series:
\[ = 2 \left( 1 - \frac{1}{n+1} \right) = 2 \left( \frac{n}{n+1} \right) \]

### Step 5: Take the limit as \( n \to \infty \)
Now we take the limit:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{a_r} = \lim_{n \to \infty} 2 \left( \frac{n}{n+1} \right) = 2 \cdot 1 = 2 \]

### Final Answer
Thus, the limit is:
\[ \boxed{2} \]