Suppose that {$a_k$} converges to 0. Prove that there exists a subsequence {$a_{p_k}$} of {$a_k$} such that $\sum_{k=1}^{\infty}a_{p_k}$ converges absolutely.
I am at a loss of how to proceed with this.
Prove that there exists a subsequence {$a_{p_k}$} of {$a_k$} such that $\sum_{k=1}^{\infty}a_{p_k}$ converges absolutely.
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real-analysis
sequences-and-series
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0Hint: since the $a_n$ go to $0$ we can find one less than $1$, then one less than $\frac 14$, then one less than $\frac 19$ and so on. (where "less" means less in absolute value). – 2017-02-27
1 Answers
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Since $a_k \to 0$, for each $n \in \mathbb N$, there is $k(n)\in \mathbb N$ such that $k \ge k(n)$ gives $$\lvert a_{k} \rvert \le \frac 1 {2^n}.$$ In particular, we can find an increasing sequence of natural numbers $p_1 \le p_2 \le p_3 \le \cdots$ such that $\lvert a_{p_k} \rvert \le 1/2^k$ for each $k$. Then $$\sum^\infty_{k = 1} \lvert a_{p_k}\rvert \le \sum^\infty_{k=1} \frac{1}{2^k} = 1$$ so we have found a subsequence whose sum converges absolutely.