Math Problem Statement

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

Solution

To prove that the series

n=11(2n1)(2n+3)\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 ϵ>0\epsilon > 0, there exists an NN such that for all m>nNm > n \geq N, the absolute difference between the partial sums is less than ϵ\epsilon:

SmSn<ϵ,\left| S_m - S_n \right| < \epsilon, where Sn=k=1n1(2k1)(2k+3)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: 1(2n1)(2n+3)=A2n1+B2n+3.\frac{1}{(2n-1)(2n+3)} = \frac{A}{2n-1} + \frac{B}{2n+3}. Solving this equation for AA and BB gives: 1=A(2n+3)+B(2n1).1 = A(2n+3) + B(2n-1). Expanding this and collecting terms leads to: 1=(A+B)2n+(3AB).1 = (A + B)2n + (3A - B). This implies two equations: A+B=0,3AB=1.A + B = 0, \quad 3A - B = 1. Solving these, we get A=18A = \frac{1}{8} and B=18B = -\frac{1}{8}.

    Thus, we can rewrite the general term as: 1(2n1)(2n+3)=18(12n112n+3).\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 nn-th partial sum SnS_n becomes: Sn=k=1n18(12k112k+3).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: Sn=18(1+1312n+112n+3).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>nm > n, consider the difference SmSnS_m - S_n: SmSn=k=n+1m1(2k1)(2k+3).S_m - S_n = \sum_{k=n+1}^{m} \frac{1}{(2k-1)(2k+3)}. Using the partial fraction decomposition, this becomes: SmSn=18(12n+1+12n+312m+112m+3).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 nn, the terms involving 12n+1,12n+3,12m+1,12m+3\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 SmSnS_m - S_n by: SmSnCn.|S_m - S_n| \leq \frac{C}{n}. Hence, as nn \to \infty, SmSn0S_m - S_n \to 0, which proves that the series is Cauchy.

Conclusion:

Since SmSnS_m - S_n can be made arbitrarily small by taking nn 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Series Convergence
Cauchy Sequence
Partial Fractions
Telescoping Series

Formulas

\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+3)}
Partial fraction decomposition: \frac{1}{(2n-1)(2n+3)} = \frac{1}{8} \left( \frac{1}{2n-1} - \frac{1}{2n+3} \right)

Theorems

Cauchy Convergence Criterion
Telescoping Series Theorem

Suitable Grade Level

Undergraduate Level