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In my Analysis course we showed that the harmonic series diverged but the alternating harmonic series converges. This got me thinking:

How many sequences $(a_n)$ where $|a_n|=1/n$ are there such that $\sum a_n$ converges?

There are quite clearly at least countably many (modify the sign of the $n$th term of the alternating series and we still have convergence), so my question is really whether there are uncountably many such sequences. I should think so but I am failing to find a good reason.

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We can freely change signs on the terms of the shape $\frac{1}{2^k}$ without affecting convergence. So there are continuum-many such sequences.

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  • $\begingroup$ Thanks, that is a clever observation. $\endgroup$ – Sonk Nov 4 '15 at 3:54
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Nov 4 '15 at 3:56
  • $\begingroup$ Out of interest can something like this always be done given a conditionally convergent series $\sum a_n?$ i.e. can we always find a subsequence $(x_n)$ of $(a_n)$ (in this case $x_n=2^{-n}$) such that $\sum x_n$ is absolutely convergent? $\endgroup$ – Sonk Nov 4 '15 at 3:59
  • $\begingroup$ Yes, because if the series $\sum a_n$ converges, then the terms have limit $0$. So we can find an $n_1$ such that $|a_{n_1}|\lt \frac{1}{2}$. And then we can find an $n_2\gt n_1$ such that $|a_{n_2}|\lt \frac{1}{2^2}$. And then we can find an $n_3$ such that $\dots$. $\endgroup$ – André Nicolas Nov 4 '15 at 4:06

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