Math Problem Statement
how can i prove the sigma sum of 1/((2n-1)(2n+3)) is cauchy?
Solution
To prove that the series
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 , there exists an such that for all , the absolute difference between the partial sums is less than :
where .
Steps to prove the series is Cauchy:
-
Rewrite the general term of the series: We start by simplifying the general term using partial fractions. Write: Solving this equation for and gives: Expanding this and collecting terms leads to: This implies two equations: Solving these, we get and .
Thus, we can rewrite the general term as:
-
Express the partial sum: The -th partial sum becomes: This is a telescoping series, meaning many terms will cancel out, leaving:
-
Bound the remainder of the series: To show that the series is Cauchy, we need to examine the tail of the series. For , consider the difference : Using the partial fraction decomposition, this becomes: For large , the terms involving become small, and we can bound the absolute value of by: Hence, as , , which proves that the series is Cauchy.
Conclusion:
Since can be made arbitrarily small by taking 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:
- How do you derive partial fraction decompositions?
- What is a telescoping series, and why do terms cancel out?
- How do we apply the Cauchy criterion for series in general?
- What are the conditions for a series to be convergent?
- 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.
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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
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