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I am interested in knowing whether both the sets described above are uncountable.

I think I managed to show that the set of binary sequences such that the series converges is uncountable. The main idea was to pick an arbitrary real $x$. Then given that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we can always construct $a_n$ such that the partial sums get arbitrarily close to $x$ and hence converge to $x$. Thus there is a surjection from all such sequences to the reals, so the set is uncountable.

I have had no luck with the set of sequences such that the series doesn't converge though. I tried the diagonal argument, but I cannot see a reason why the 'new sequence' would necessarily yield a divergent series. Any ideas?

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  • $\begingroup$ What is the sequence that, for example, gives a series that converges to $\pi$? $\endgroup$ Dec 29, 2013 at 17:06
  • $\begingroup$ @FlybyNight I doubt a general term can be expressed explicitly - the point is that such a sequence exists. $\endgroup$
    – WPL
    Dec 29, 2013 at 17:08
  • $\begingroup$ How can we be sure it exists? $\endgroup$ Dec 29, 2013 at 17:21
  • $\begingroup$ @Phira Beautifully simple. Thank you. $\endgroup$ Dec 29, 2013 at 17:44
  • $\begingroup$ You start to add terms until you overshoot $\pi$, then you subtract until you are below $\pi$, then you add again and so on. The divergence of the harmonic series guarantees that you can cross $\pi$ again and again and the monotonous convergence of the sequence $\frac1n$ to zero guarantees that the deviation from $\pi$ gets smaller and smaller. Hence, it converges. $\endgroup$
    – Phira
    Dec 29, 2013 at 17:48

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The set of divergent series is also uncountable. Choose $a_{2k}=1$, so that the sub-sequence of even terms already tends to infinity. Now, like in your part of the proof, let the odd terms tend to a real numbers $x$. The combination still diverges to infinity, so we have found uncountably many divergent series.

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  • $\begingroup$ Doh!... Thanks! $\endgroup$
    – WPL
    Dec 29, 2013 at 17:07

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