We subtract 1 from both sides of the equation:
\[ 9 = 4k^2 + 8k^3 + 13k^4 + 19k^5 + \dots \]
Next, we multiply the entire equation by \( k \) and reorganize terms:
\[ 9k = 4k^3 + 8k^4 + 13k^5 + \dots \]
The series can be expressed as:
\[ S = 9k - 4k^2 - 8k^3 - 13k^4 + \dots \]
We now simplify the series and make use of the infinite geometric series form starting from \( k^3 \):
\[ S = 4k + 4k^2 + 5k^3 + 6k^4 + \dots \]
The series represents a geometric progression, and we solve for \( S \) by simplifying:
\[ S = 4k + 1 + k^3 + \dots \]
We equate the series and solve for \( k \). After simplifying and solving:
\[ k = 2 \]
The value of \( k \) is \( 2 \).
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :