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Show that sequence $x_n$ is absolutely summable if and only if for every $\epsilon>0$ there is an $N\in\mathbb{N}$ such that for every finite subset $F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$, $\displaystyle{\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$.

$\textbf{Proof:}$ Let the sequence $x_n$ be absolutely summable so the series $\displaystyle{\sum_{n=1}^\infty |x_n|}$ is convergent. Therefore, by the Cauchy criterion, for every $\epsilon > 0$ there exists a natural number $N$ such that $$ \bigl\lvert\lvert x_n \rvert +\lvert x_{n+1} \rvert + \dots + \lvert x_{n+p} \rvert \bigr\rvert \\ < \epsilon.$$ Thereby, implying $\displaystyle{\sum_{i=n}^{n+p} |x_i| < \epsilon}$ for all $n\geq N$ and $p\geq 1.$ This comes directly from the Cauchy Criterion of Sequence Convergence and has an alternate form $$ \bigl\lvert \sum_{i=n}^m \lvert x_i \rvert \bigr\rvert \\ < \epsilon$$ for all $n\geq N$ and $m>n$.

Now, consider any finite subset $F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$ then let min$F = n_1$ and max$F=m_1$ where they exist since F is a finite subset of $\mathbb{N}$ and $n_1\geq N$ and $m_1 \geq n_1$. Therefore, as the Cauchy criterion we have, $$ \bigl\lvert \sum_{i=n_1}^{m_1} \lvert x_i \rvert \bigr\rvert \\ < \epsilon$$ as $n_1\geq N$ implies $\displaystyle{\sum_{i=n_1}^{m_1} |x_i| < \epsilon}.$. Then, $\displaystyle{\sum_{n\in F} |x_n| \leq \sum_{i=n_1}^{m_1} |x_i| < \epsilon}$ implies $\displaystyle{|\sum_{n\in F} x_n| \leq \sum_{n\in F} |x_n| < \epsilon}$. Therefore, this show that there exists an $N$ such that for every finite subset $F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$, $\displaystyle{\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$. Clearly, we have a generalization of Cauchy Criterion and the condition came as Cauchy Criterion when $F$ has the form $F=\{n, n+1, \dots, m\}$.

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You proved one direction: If $\sum_{n=1}^\infty x_n$ converges absolutely then for every $\epsilon>0$ there is an $N\in\mathbb{N}$ such that for every finite subset $F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$, ${\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$.


For the opposite direction assume that $\sum_{n=1}^\infty x_n$ does not converge absolutely, i.e. that $\sum_{n=1}^\infty |x_n|$ diverges. From $$ \sum_{n=1}^k |x_n| = \sum_{n=1}^k x_n^+ + \sum_{n=1}^k x_n^- $$ with $x_n^+ = \max(x_n, 0)$, $x_n^- = \max(-x_n, 0)$ it follows that at least one of the series $\sum_{n=1}^\infty x_n^+$, $\sum_{n=1}^\infty x_n^-$ diverges.

Without loss of generality assume that $\sum_{n=1}^\infty x_n^+$ diverges, i.e. $$ \lim_{k \to \infty} \sum_{n=1}^k x_n^+ = + \infty \, . $$ It follows that for every $N \in \Bbb N$ $$ \lim_{k \to \infty} \sum_{n=N}^k x_n^+ = + \infty $$ and that implies $$ \bigg|\sum_{n\in \{ N, \ldots, m\}} x_n\bigg| = \sum_{n=N}^m x_n^+ > 1 $$ for some $m > N$.

So we have shown that if $\sum_{n=1}^\infty x_n$ does not converge absolutely then for $\epsilon = 1$ there is no $N \in \Bbb N$ such that ${\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$ for every finite subset $F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$. That proves the other direction by contrapositive.

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  • $\begingroup$ Thank you for the explanation! $\endgroup$ Commented Feb 11, 2020 at 9:39

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