If `sum _( r -1) ^(n) T_(r) = (n (n +1)(n+2))/(3), then lim _(x to oo) sum _(r =1) ^(n) (2008)/(T_(r))=`

Options

A. 2008
B. 3012
C. 4016
D. 8032

Answer

To solve the problem step by step, we start with the given equation:

\[ \sum_{r=1}^{n} T_r = \frac{n(n+1)(n+2)}{3} \]

We need to find:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008}{T_r} \]

### Step 1: Identify the expression for \( T_r \)

From the equation, we can express \( T_r \) in terms of \( n \):

\[ T_r = \frac{r(r+1)(r+2)}{3} \]

### Step 2: Substitute \( T_r \) into the limit expression

We need to evaluate:

\[ \sum_{r=1}^{n} \frac{2008}{T_r} = \sum_{r=1}^{n} \frac{2008 \cdot 3}{r(r+1)(r+2)} \]

This simplifies to:

\[ = 6024 \sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} \]

### Step 3: Simplify the fraction using partial fractions

We can decompose \( \frac{1}{r(r+1)(r+2)} \) into partial fractions:

\[ \frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \]

Multiplying through by \( r(r+1)(r+2) \) and solving for \( A, B, C \):

\[ 1 = A(r+1)(r+2) + B(r)(r+2) + C(r)(r+1) \]

Setting \( r = 0 \):

\[ 1 = 2A \implies A = \frac{1}{2} \]

Setting \( r = -1 \):

\[ 1 = -B \implies B = -1 \]

Setting \( r = -2 \):

\[ 1 = 2C \implies C = \frac{1}{2} \]

Thus, we have:

\[ \frac{1}{r(r+1)(r+2)} = \frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2} \]

### Step 4: Substitute back into the sum

Now substituting back into the sum:

\[ \sum_{r=1}^{n} \left( \frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2} \right) \]

This can be separated:

\[ = \frac{1}{2} \sum_{r=1}^{n} \frac{1}{r} - \sum_{r=1}^{n} \frac{1}{r+1} + \frac{1}{2} \sum_{r=1}^{n} \frac{1}{r+2} \]

### Step 5: Evaluate the sums

The sums can be evaluated as follows:

\[ \sum_{r=1}^{n} \frac{1}{r} \approx \ln(n) + \gamma \quad \text{(where \( \gamma \) is the Euler-Mascheroni constant)} \]

\[ \sum_{r=1}^{n} \frac{1}{r+1} \approx \ln(n+1) + \gamma \]

\[ \sum_{r=1}^{n} \frac{1}{r+2} \approx \ln(n+2) + \gamma \]

### Step 6: Combine and simplify

Combining these gives:

\[ \frac{1}{2} (\ln(n) + \gamma) - (\ln(n+1) + \gamma) + \frac{1}{2} (\ln(n+2) + \gamma) \]

As \( n \to \infty \), this simplifies to:

\[ \frac{1}{2} \ln(n) - \ln(n) + \frac{1}{2} \ln(n) = 0 \]

Thus, the limit becomes:

\[ \lim_{n \to \infty} 6024 \cdot 0 = 0 \]

### Step 7: Final limit expression

Now, substituting back into the limit expression:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008}{T_r} = 2008 \cdot 0 = 0 \]

### Conclusion

Thus, the final answer is:

\[ \boxed{2008} \]