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We are asked to analyze a sequence and its corresponding partial sum sequence, then determine if it converges and find the limit if it does.

### Problem Setup:
Given the sequence \(\{a_n\}\), where:
\[
a_n = \frac{1}{4n^2 - 1}
\]
Define a sequence of partial sums \(\{s_n\}\) by:
\[
s_n = a_1 + a_2 + \cdots + a_n
\]
We need to determine whether \(\{s_n\}\) converges, and if so, find its limit.

### Step-by-Step Solution:

#### Step 1: Understanding the Sequence \(\{a_n\}\)
The sequence \(\{a_n\}\) is defined as:
\[
a_n = \frac{1}{4n^2 - 1}
\]
To simplify, notice that \(4n^2 - 1\) can be factored:
\[
a_n = \frac{1}{(2n - 1)(2n + 1)}
\]

#### Step 2: Express \(a_n\) as a Telescoping Series
The goal here is to express \(a_n\) in a form that allows for cancellation when summed. We use partial fractions:
\[
\frac{1}{(2n - 1)(2n + 1)} = \frac{A}{2n - 1} + \frac{B}{2n + 1}
\]
Multiplying both sides by \((2n - 1)(2n + 1)\), we get:
\[
1 = A(2n + 1) + B(2n - 1)
\]
This simplifies to:
\[
1 = A(2n) + A + B(2n) - B
\]
\[
1 = (A + B)2n + (A - B)
\]
Equating coefficients, we get two equations:
\[
A + B = 0 \quad \text{(1)}
\]
\[
A - B = 1 \quad \text{(2)}
\]
From equation (1), \(A = -B\). Substituting into equation (2), we get:
\[
A - (-A) = 1 \quad \Rightarrow \quad 2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}
\]
Since \(A = \frac{1}{2}\) and \(A = -B\), we have \(B = -\frac{1}{2}\).

Thus, the partial fraction decomposition is:
\[
a_n = \frac{1}{2}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)
\]

#### Step 3: Summing the Series \(\{a_n\}\)
The partial sum \(s_n\) becomes:
\[
s_n = \sum_{k=1}^n a_k = \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right)
\]
This is a telescoping series, meaning that most terms cancel out. Writing out the first few terms of the sum, we see:
\[
s_n = \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \right)
\]
All intermediate terms cancel, leaving us with:
\[
s_n = \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right)
\]

#### Step 4: Determining Convergence
As \(n \to \infty\), the term \(\frac{1}{2n+1} \to 0\). Thus, the sum converges to:
\[
\lim_{n \to \infty} s_n = \frac{1}{2} \times 1 = \frac{1}{2}
\]

### Conclusion:
The sequence of partial sums \(\{s_n\}\) converges, and the limit is:
\[
\lim_{n \to \infty} s_n = \frac{1}{2}
\]

Would you like to see more details on any specific part of the solution? 

Here are 5 related questions:
1. How do we determine whether a series converges using telescoping?
2. Why does the cancellation of terms occur in a telescoping series?
3. What is the significance of partial fraction decomposition in this problem?
4. How do we prove the convergence of a sequence rigorously?
5. Can this method be applied to any other sequences of a similar form?

**Tip:** Telescoping series is a very efficient method for finding sums when terms cancel out. Always look for patterns that allow for simplification!

Consider the sequence {a_n}, where a_n = 1 / (4n^2 - 1). Define the sequence {s_n} by s_n = a_1 + a_2 + ... + a_n. Determine whether or not {s_n} converges. If so, find the limit.

Explore the convergence of a sequence {s_n} defined by partial sums of a_n = 1 / (4n^2 - 1). Using telescoping series, we determine if the sequence converges and find the limit. This step-by-step solution covers sequences, series, and limits, ideal for university-level students studying Calculus II or sequences and series.

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