If `sum_(r=1)^(n) T_(r)=(n(n+1)(n+2)(n+3))/(8)`, then
`lim_(ntooo) sum_(r=1)^(n) (1)/(T_(r))=`

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(8)`

Explanation

To solve the problem, we need to find the limit:

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

where it is given that:

\[ \sum_{r=1}^{n} T_r = \frac{n(n+1)(n+2)(n+3)}{8} \]

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

We know that:

\[ T_r = S_r - S_{r-1} \]

where \( S_n = \sum_{r=1}^{n} T_r \). Thus, we can express \( T_r \) as:

\[ T_r = \frac{r(r+1)(r+2)(r+3)}{8} - \frac{(r-1)r(r+1)(r+2)}{8} \]

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

Calculating \( T_r \):

\[ T_r = \frac{1}{8} \left[ r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2) \right] \]

Expanding both terms:

1. \( r(r+1)(r+2)(r+3) = r^4 + 6r^3 + 11r^2 + 6r \)
2. \( (r-1)r(r+1)(r+2) = r^4 + 3r^3 - 2r^2 - 2r \)

Subtracting the second from the first:

\[ T_r = \frac{1}{8} \left[ (r^4 + 6r^3 + 11r^2 + 6r) - (r^4 + 3r^3 - 2r^2 - 2r) \right] \]

This simplifies to:

\[ T_r = \frac{1}{8} \left[ 3r^3 + 13r^2 + 8r \right] \]

### Step 3: Find \( \frac{1}{T_r} \)

Now we need to find \( \frac{1}{T_r} \):

\[ \frac{1}{T_r} = \frac{8}{3r^3 + 13r^2 + 8r} \]

### Step 4: Sum \( \frac{1}{T_r} \) from \( r=1 \) to \( n \)

We need to compute:

\[ \sum_{r=1}^{n} \frac{1}{T_r} = \sum_{r=1}^{n} \frac{8}{3r^3 + 13r^2 + 8r} \]

### Step 5: Analyze the limit as \( n \to \infty \)

As \( n \to \infty \), we can approximate \( T_r \):

\[ T_r \sim \frac{3r^3}{8} \quad \text{(for large } r\text{)} \]

Thus,

\[ \frac{1}{T_r} \sim \frac{8}{3r^3} \]

### Step 6: Evaluate the sum

The sum can be approximated as:

\[ \sum_{r=1}^{n} \frac{8}{3r^3} \sim \frac{8}{3} \cdot \int_{1}^{n} \frac{1}{x^3} \, dx \]

Calculating the integral:

\[ \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} \]

Thus,

\[ \int_{1}^{n} \frac{1}{x^3} \, dx = -\frac{1}{2n^2} + \frac{1}{2} \]

So,

\[ \sum_{r=1}^{n} \frac{8}{3r^3} \sim \frac{8}{3} \left( \frac{1}{2} - \frac{1}{2n^2} \right) \]

### Step 7: Take the limit

Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{T_r} = \frac{8}{3} \cdot \frac{1}{2} = \frac{4}{3} \]

Thus, the final answer is:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{T_r} = \frac{4}{3} \]