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Let $(a_n)$ be a sequence such that $\liminf |a_n| = 0$. Then there exists a subsequence $(a_{n_k})$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.

any thoughts?

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Since the $\liminf|a_n| = 0$, you can choose a subsequence satisfying, say, $|a_{n_k}| \leq 2^{-k}$. –  Austin Mohr Oct 17 '12 at 7:33

1 Answer 1

up vote 2 down vote accepted

Here’s an outline, leaving a lot of the detail for you.

You want to find a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ of $\langle a_n:n\in\Bbb Z^+\rangle$ that approaches $0$ so quickly that $\sum_{k\ge 1}a_{n_k}$ converges. Start by using the fact that $\liminf_na_n=0$ to infer that $\langle a_n:n\in\Bbb Z^+\rangle$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ converging to $0$. This implies that for each $n\in\Bbb Z^+$ there is a $k\in\Bbb Z^+$ such that $|a_{n_k}|<\frac1{2^n}$. Use this to construct an increasing sequence $\langle k(i):i\in\Bbb Z^+\rangle$ such that $$|a_{n_{k(i)}}|<\frac1{2^i}$$ for each $i\in\Bbb Z^+$; $\langle a_{n_{k(i)}}:i\in\Bbb Z^+\rangle$ is a subsequence of $\langle a_n:n\in\Bbb Z^+\rangle$.

Finally, explain why $\displaystyle\sum_{i\ge 1}a_{n_{k(i)}}$ converges. (Indeed, you should be able to explain why it converges to some number in the interval $[-1,1]$.)

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