Math Problem Statement

把级数

112α+13α14α+15α16α+(0<α<1)1 - \frac{1}{2^\alpha} + \frac{1}{3^\alpha} - \frac{1}{4^\alpha} + \frac{1}{5^\alpha} - \frac{1}{6^\alpha} + \cdots \quad (0 < \alpha < 1)

的项重新排列如下:先依次取 $p$ 个正项,接着依次取 $q$个负 项,再接着依次取 $p$ 个正项,如此继续下去。证明:所得的新级数收敛的充分必要条件为 $p = q$;当 $p > q$ 时,新级数发散到 $+\infty$;当 $p < q$ 时,新级数发散到 $-\infty$。

第1小问证明所得的新级数收敛的充分必要条件为 $p = q$。我认为充分性可以这样证明,首先p=q=1的时候,刚好新级数就是原级数,根据莱布尼茨定理是条件收敛的。接下来用数学归纳法,假设p= q=k-1(k为大于1的正整数)时,新级数收敛。下证p=q=k的时候。 你认为我的思路对吗?

Solution

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Math Problem Analysis

Mathematical Concepts

Series and Sequences
Conditional Convergence
Rearrangement of Series

Formulas

Alternating series: $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ where $a_n > 0$

Theorems

Leibniz's Alternating Series Test
Convergence of Rearranged Series

Suitable Grade Level

Undergraduate Mathematics