The given sequence is:
$x_n = \frac{1}{5} + \frac{1}{21} + \cdots + \frac{1}{(2n-1)(2n+3)}$

To find the limit of this sequence ($\lim_{n \to \infty} x_n$), let's examine it term by term.

Step 1: General term representation

Each term of the sequence can be written as:
$a_k = \frac{1}{(2k-1)(2k+3)}.$

Step 2: Simplify the general term using partial fraction decomposition

We decompose the term: $\frac{1}{(2k-1)(2k+3)} = \frac{A}{2k-1} + \frac{B}{2k+3}.$ Solving for $A$ and $B$, we get: $1 = A(2k+3) + B(2k-1).$ Substituting values:

• Coefficient of $2k$: $2A + 2B = 0 \implies A = -B$,
• Constant term: $3A - B = 1 \implies 3(-B) - B = 1 \implies -4B = 1 \implies B = -\frac{1}{4}, A = \frac{1}{4}$.

Thus: $\frac{1}{(2k-1)(2k+3)} = \frac{1/4}{2k-1} - \frac{1/4}{2k+3}.$

Step 3: Rewrite the sequence as a telescoping series

The sequence becomes: $x_n = \sum_{k=1}^n \left( \frac{1}{4(2k-1)} - \frac{1}{4(2k+3)} \right).$ Expanding this telescoping series: $x_n = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{7} + \frac{1}{3} - \frac{1}{9} + \frac{1}{5} - \cdots - \frac{1}{2n+3} \right).$

Step 4: Evaluate the limit as $n \to \infty$

As $n \to \infty$, most terms cancel out in the telescoping sum, leaving: $x_n \to \frac{1}{4} \left( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} - \lim_{n \to \infty} \frac{1}{2n+3} \right).$

Since $\frac{1}{2n+3} \to 0$ as $n \to \infty$, the limit becomes: $\lim_{n \to \infty} x_n = \frac{1}{4} \left( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots \right).$

This is a convergent series of reciprocals of odd integers: $\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots = \text{a known logarithmic divergence result}.$

Let me know if you'd like clarification or a step!