MY-ASSIGNMENTEXPERT™可以为您提供catalog.upp.pitt.ed MATH0450 Mathematical Analysis数学分析课程的代写代考和辅导服务！
MATH 0450课程简介
Minimum Credits: 4
Maximum Credits: 4
This course is intended as a first course in mathematical analysis for highly motivated students. Topics will include sets and functions, number systems, topology of Euclidean spaces, limits, continuity, and the main theorems of elementary calculus.
Prerequisites
Academic Career: UGRD
Course Component: Lecture
Grade Component: LG/SNC Elective Basis
MATH0450 Mathematical Analysis HELP（EXAM HELP， ONLINE TUTOR）
Fix $\epsilon>0$. Since $\sum_{i=1}^{\infty} a_i$ converges, there exists $N \in \mathbb{N}$ such that for any $m, n>N$, $\sum_{i=m}^n a_i<\epsilon^{\frac{1}{3}}$. Then $\sum_{i=m}^n a_i^3<\left(\sum_{i=m}^n a_i\right)^3<\epsilon$. Hence, $\sum_{i=1}^{\infty} a_i^3$ converges.
The proof is not correct. The statement that $\sum_{i=m}^n a_i^3<\left(\sum_{i=m}^n a_i\right)^3$ is not true in general, as it depends on the signs of the $a_i$’s. In fact, if some of the $a_i$’s are negative, the inequality is reversed, and the proof fails.
To prove that $\sum_{i=1}^\infty a_i^3$ converges, we can use the Cauchy-Schwarz inequality as follows:
\begin{align*} \sum_{i=1}^n |a_i^3| &= \sum_{i=1}^n |a_i| |a_i|^2 \ &\leq \left(\sum_{i=1}^n |a_i|\right)\left(\sum_{i=1}^n |a_i|^2\right) \ &\leq \left(\sum_{i=1}^\infty |a_i|\right)\left(\sum_{i=1}^\infty |a_i|^2\right) \quad \text{(by the Cauchy-Schwarz inequality)}\ &< \infty \quad \text{(since $\sum_{i=1}^\infty |a_i|$ and $\sum_{i=1}^\infty |a_i|^2$ converge)} \end{align*}
Therefore, by the comparison test, $\sum_{i=1}^\infty a_i^3$ converges absolutely, and hence converges.
Counter-example: $\sum_{n=1}^{\infty} \frac{1}{n^3}$ converges but $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges. (See Textbook Example 3.7.6 for a proof of the divergence.)
I apologize for the mistake in my previous response. You are correct that the series $\sum_{n=1}^\infty a_n^3$ may converge even if the series $\sum_{n=1}^\infty a_n$ diverges. The example you gave, $\sum_{n=1}^\infty \frac{1}{n^3}$, is a good counterexample. Thank you for pointing this out.
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