$$ \text{If } \sum_{r=1}^n T_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64} \text{ then } \lim_{n \rightarrow \infty} \sum_{r=1}^n\left( \frac {1}{T_r}\right) \text{is equal to:} $$

Given that the sum of the first $$n$$ terms is:

$$\sum_{r=1}^n T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$$

To find the general term $$T_n$$, use the relation $$T_n = S_n - S_{n-1}$$ for $$n \geq 2$$, where $$S_n$$ is the sum up to $$n$$ terms.

First, express $$S_n$$ and $$S_{n-1}$$:

$$S_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$$

$$S_{n-1} = \frac{(2(n-1)-1)(2(n-1)+1)(2(n-1)+3)(2(n-1)+5)}{64} = \frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}$$

Now, compute $$T_n = S_n - S_{n-1}$$:

$$T_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64} - \frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}$$

Factor out the common terms:

$$T_n = \frac{(2n-1)(2n+1)(2n+3)}{64} \left[ (2n+5) - (2n-3) \right]$$

Simplify the expression inside the brackets:

$$(2n+5) - (2n-3) = 8$$

So,

$$T_n = \frac{(2n-1)(2n+1)(2n+3)}{64} \times 8 = \frac{8}{64} \times (2n-1)(2n+1)(2n+3) = \frac{1}{8} (2n-1)(2n+1)(2n+3)$$

Thus,

$$T_n = \frac{(2n-1)(2n+1)(2n+3)}{8}$$

Now, find $$\frac{1}{T_n}$$:

$$\frac{1}{T_n} = \frac{8}{(2n-1)(2n+1)(2n+3)}$$

Decompose this into partial fractions. Set:

$$\frac{8}{(2n-1)(2n+1)(2n+3)} = \frac{A}{2n-1} + \frac{B}{2n+1} + \frac{C}{2n+3}$$

Multiply both sides by $$(2n-1)(2n+1)(2n+3)$$:

$$8 = A(2n+1)(2n+3) + B(2n-1)(2n+3) + C(2n-1)(2n+1)$$

Expand the right-hand side:

$$A(2n+1)(2n+3) = A(4n^2 + 8n + 3)$$

$$B(2n-1)(2n+3) = B(4n^2 + 4n - 3)$$

$$C(2n-1)(2n+1) = C(4n^2 - 1)$$

So,

$$8 = (4A + 4B + 4C)n^2 + (8A + 4B)n + (3A - 3B - C)$$

Equate coefficients of like powers of $$n$$:

For $$n^2$$: $$4A + 4B + 4C = 0$$ ⇒ $$A + B + C = 0$$ $$-(1)$$

For $$n$$: $$8A + 4B = 0$$ ⇒ $$2A + B = 0$$ $$-(2)$$

Constant term: $$3A - 3B - C = 8$$ $$-(3)$$

Solve the system of equations. From equation (2):

$$B = -2A$$

Substitute into equation (1):

$$A + (-2A) + C = 0 ⇒ -A + C = 0 ⇒ C = A$$

Substitute $$B = -2A$$ and $$C = A$$ into equation (3):

$$3A - 3(-2A) - A = 8 ⇒ 3A + 6A - A = 8 ⇒ 8A = 8 ⇒ A = 1$$

Then, $$B = -2(1) = -2$$ and $$C = 1$$.

So,

$$\frac{8}{(2n-1)(2n+1)(2n+3)} = \frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}$$

Therefore,

$$\frac{1}{T_n} = \frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}$$

Now, compute the sum:

$$\sum_{r=1}^n \frac{1}{T_r} = \sum_{r=1}^n \left( \frac{1}{2r-1} - \frac{2}{2r+1} + \frac{1}{2r+3} \right)$$

This is a telescoping series. Write the sum explicitly:

$$\sum_{r=1}^n \frac{1}{2r-1} - 2 \sum_{r=1}^n \frac{1}{2r+1} + \sum_{r=1}^n \frac{1}{2r+3}$$

Adjust the indices for alignment:

The first sum: $$\sum_{r=1}^n \frac{1}{2r-1} = \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1}$$

The second sum: $$\sum_{r=1}^n \frac{1}{2r+1} = \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n+1}$$

The third sum: $$\sum_{r=1}^n \frac{1}{2r+3} = \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{2n+3}$$

Combine the sums:

$$S_n = \left( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} \right) - 2 \left( \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n+1} \right) + \left( \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{2n+3} \right)$$

Group terms by denominator:

• Denominator 1: only in the first sum, coefficient $$+1$$ → term $$+\frac{1}{1}$$
• Denominator 3: in first sum ($$+1$$) and second sum ($$-2$$), coefficient $$1 - 2 = -1$$ → term $$-\frac{1}{3}$$
• Denominators 5, 7, ..., $$2n-1$$: in first sum ($$+1$$), second sum ($$-2$$), and third sum ($$+1$$), coefficient $$1 - 2 + 1 = 0$$
• Denominator $$2n+1$$: in second sum ($$-2$$) and third sum ($$+1$$), coefficient $$-2 + 1 = -1$$ → term $$-\frac{1}{2n+1}$$
• Denominator $$2n+3$$: only in third sum, coefficient $$+1$$ → term $$+\frac{1}{2n+3}$$

Thus, the sum simplifies to:

$$S_n = \frac{1}{1} - \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+3}$$

Simplify:

$$S_n = \left(1 - \frac{1}{3}\right) + \left(-\frac{1}{2n+1} + \frac{1}{2n+3}\right) = \frac{2}{3} + \left(\frac{1}{2n+3} - \frac{1}{2n+1}\right)$$

Compute the difference:

$$\frac{1}{2n+3} - \frac{1}{2n+1} = \frac{(2n+1) - (2n+3)}{(2n+1)(2n+3)} = \frac{-2}{(2n+1)(2n+3)}$$

So,

$$S_n = \frac{2}{3} - \frac{2}{(2n+1)(2n+3)}$$

Now, evaluate the limit as $$n \to \infty$$:

$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[ \frac{2}{3} - \frac{2}{(2n+1)(2n+3)} \right]$$

As $$n \to \infty$$, $$\frac{2}{(2n+1)(2n+3)} \to 0$$, so:

$$\lim_{n \to \infty} S_n = \frac{2}{3}$$

Thus, the limit is $$\frac{2}{3}$$, which corresponds to option B.