how can i prove the sigma sum of 1/((2n-1)(2n+3)) is cauchy?

To prove that the series

\[
\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+3)}
\]

is a **Cauchy series**, we need to verify that the sequence of partial sums of the series converges. Specifically, we must show that for any given \(\epsilon > 0\), there exists an \(N\) such that for all \(m > n \geq N\), the absolute difference between the partial sums is less than \(\epsilon\):

\[
\left| S_m - S_n \right| < \epsilon,
\]
where \(S_n = \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+3)}\).

### Steps to prove the series is Cauchy:

1. **Rewrite the general term of the series**:
   We start by simplifying the general term using partial fractions. Write:
   \[
   \frac{1}{(2n-1)(2n+3)} = \frac{A}{2n-1} + \frac{B}{2n+3}.
   \]
   Solving this equation for \(A\) and \(B\) gives:
   \[
   1 = A(2n+3) + B(2n-1).
   \]
   Expanding this and collecting terms leads to:
   \[
   1 = (A + B)2n + (3A - B).
   \]
   This implies two equations:
   \[
   A + B = 0, \quad 3A - B = 1.
   \]
   Solving these, we get \(A = \frac{1}{8}\) and \(B = -\frac{1}{8}\).

   Thus, we can rewrite the general term as:
   \[
   \frac{1}{(2n-1)(2n+3)} = \frac{1}{8} \left( \frac{1}{2n-1} - \frac{1}{2n+3} \right).
   \]

2. **Express the partial sum**:
   The \(n\)-th partial sum \(S_n\) becomes:
   \[
   S_n = \sum_{k=1}^{n} \frac{1}{8} \left( \frac{1}{2k-1} - \frac{1}{2k+3} \right).
   \]
   This is a telescoping series, meaning many terms will cancel out, leaving:
   \[
   S_n = \frac{1}{8} \left( 1 + \frac{1}{3} - \frac{1}{2n+1} - \frac{1}{2n+3} \right).
   \]

3. **Bound the remainder of the series**:
   To show that the series is Cauchy, we need to examine the tail of the series. For \(m > n\), consider the difference \(S_m - S_n\):
   \[
   S_m - S_n = \sum_{k=n+1}^{m} \frac{1}{(2k-1)(2k+3)}.
   \]
   Using the partial fraction decomposition, this becomes:
   \[
   S_m - S_n = \frac{1}{8} \left( \frac{1}{2n+1} + \frac{1}{2n+3} - \frac{1}{2m+1} - \frac{1}{2m+3} \right).
   \]
   For large \(n\), the terms involving \( \frac{1}{2n+1}, \frac{1}{2n+3}, \frac{1}{2m+1}, \frac{1}{2m+3} \) become small, and we can bound the absolute value of \(S_m - S_n\) by:
   \[
   |S_m - S_n| \leq \frac{C}{n}.
   \]
   Hence, as \(n \to \infty\), \(S_m - S_n \to 0\), which proves that the series is Cauchy.

### Conclusion:
Since \(S_m - S_n\) can be made arbitrarily small by taking \(n\) large enough, the series satisfies the Cauchy criterion for convergence. Therefore, the series is convergent.

---

Would you like a deeper explanation of any step, or have further questions?

### Related Questions:
1. How do you derive partial fraction decompositions?
2. What is a telescoping series, and why do terms cancel out?
3. How do we apply the Cauchy criterion for series in general?
4. What are the conditions for a series to be convergent?
5. How can I estimate the remainder of a convergent series?

### Tip:
A telescoping series simplifies because consecutive terms cancel each other out, making it easier to find the partial sum.

Explore the step-by-step process to prove that the series \sum_{n=1}^{\infty} 1/((2n-1)(2n+3)) is a Cauchy series. This detailed guide covers key mathematical concepts such as series convergence, partial fractions, and the Cauchy criterion, suitable for undergraduate students.

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