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Let $\{a_n\}$ be a sequence of reals such that $a_n=1$ or $a_n=-1$.

Let $A=\{n\in \mathbb{N}|a_n=1\}$ and $B=\{n\in \mathbb{N}|a_n=-1\}$.

Suppose $A$ is equipotent with $B$. That is, $|A|=|B|=\aleph_0$.

Here, how do i prove that partial sum of $\sum a_n$ form a bounded sequence?

Or is it false?

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    It is false: take $A = \{3k|k\in \mathbb N\}$ and $B = \{3k+1,3k+2|k\in \Bbb N\}$, then clearly $S_{3(n+1)} = S_{3n}-1$ which means that the sequence $(S_n)_{n\in \Bbb N}$ of partial sums is unbounded.2012-08-29
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    Another assertion is the following: suppose, $(S_n)_{n\in \Bbb N}$ stays bounded. Then it has a convergent subsequence - i.e. the sum $\sum\limits_{k=1}^\infty a_{n_k}$. But then it would mean that $a_{n_k}\to 0$ which can't happen as $a_{n_k} = \pm 1$.2012-08-29
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    Thank you. Does this mean there is no wider generalization than '$a_{2k} ≧ 0$ and $a_{2k+1} < 0$' for convergence of alternating series?2012-08-29
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    even in that case you need $a_k\to 0$ - don't forget this important fact. But I don't think that if you add this condition in OP, you'll obtain the convergence: even though $A$ is infinite, it can have form e.g. $\{2^n|n\in \Bbb N\}$ and if you take $|a_n| = \frac1n$ then still series diverge, and hence are unbounded.2012-08-29
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    @Ilya I think your second assertion is false. Subsequence of $(S_n)_{n\in \mathbb{N}}$ would be of the form $\sum\limits_{i=1}^{n_k} a_i$, not $\sum\limits_{i=1}^k a_{n_i}$2012-08-29
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    You're right, thanks.2012-08-29
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    Say, $a_{2i}=1$ and $a_{2i+1}=-1$. Then $\sum\limits_{i=1}^{2k} a_i$ is $0, \forall k\in \mathbb{N}$. Thus is a convergent subsequence, but this doesn't mean that $a_{n_k} →0$2012-08-29
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    Katlus: One should ask that $A$ has asymptotic density equal to 1/2 (and that so has $B$) but even this necessary condition is not sufficient.2012-08-29

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