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Suppose $\sum_{n=1}^\infty \frac{1}{n} = S$ where $S$ is finite. Then

$$S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} > \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S$$

which is a contradiction. Is this valid?

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It is great...if you can prove that changing the order of summation doesn't alter the sum (which is true if you assume the harmonic series converges by absolute convergence, but perhaps you can't use this here) . BTW, your acceptation rate is going to begin diverging... –  DonAntonio Aug 10 '12 at 2:33
    
You rearrange infinitely many terms and you need to prove that this is valid in the sense that it preserves the value of the sum. (This crucially depends on the fact that all of the terms involved are positive; if they aren't, see en.wikipedia.org/wiki/Riemann_series_theorem ). –  Qiaochu Yuan Aug 10 '12 at 2:33
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Reminds me of math.stackexchange.com/questions/29450/… –  sdcvvc Aug 10 '12 at 2:38
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The inequality should be reversed. –  lhf Aug 10 '12 at 2:46
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Rearrangement of a series is OK for nonnegative series, even if they diverge. –  GEdgar Aug 10 '12 at 3:17

1 Answer 1

up vote 5 down vote accepted

One way to express your argument more formally is by considering the partial sums

$$S_{2m}=\sum_{n=1}^{2m}\frac1n=\sum_{n=1}^m\left(\frac1{2n-1}+\frac1{2n}\right)\;.$$

You can bound these partial sums as you did to obtain

$$ S_{2m}\gt\sum_{n=1}^m\left(\frac1{2n}+\frac1{2n}\right)=\sum_{n=1}^m\frac1n=S_m\;.$$

If the series were to converge, the sequences $S_{2m}$ and $S_m$ would have to converge to a common limit, so their difference would have to converge to $0$ for $m\to\infty$, whereas in fact the gap between them increases with $m$.

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