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Let's consider the alternating harmonic series $S_n = 1-\frac12 + \frac13 - \cdots + (-1)^n\frac1n$.

By rearranging its terms, we get $S_n = (1-\frac12)-\frac14 + (\frac13-\frac16)-\frac18 + (\frac15-\frac1{10})-\cdots$.

This equals to $S_n = \frac12-\frac14 + \frac16-\frac18 + \frac1{10}-\cdots$.

By extracting $\frac12$ as common factor, we get:

$S_n = \frac12(1-\frac12 + \frac13-\frac14 + \cdots)$. So in essence, $S_n = \frac12 S_n$, therefore $1=\frac12$.

I have read the wikipedia article about Riemann series and roughly my understanding is that if the series converges, we can rearrange the terms and get any other number, or even to diverge. What could be an acceptable explanation of the paradox? Obviously 1 does not equal $\frac12$! In which of the above steps lies the error?

user1176409
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Marius Stephant
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    To arrange terms, it must be absolutely convergent. – hamam_Abdallah Mar 05 '18 at 19:45
  • Further, a conditionally convergent series can be rearranged to converge to any real $x$; every rearrangement of an absolutely convergent series converges to the same value. – Nap D. Lover Mar 05 '18 at 19:47
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    It seems you are only considering partial sums $S_n$ hence the considerations of (absolute) convergence and rearrangement are offtopic. However, if you make the "$\cdots$" in your post explicit, you will see that one does not reach "in essence" $S_n=\frac12S_n$. – Did Mar 05 '18 at 19:47
  • @Did: I thought so, but which are the terms that I've left out? – Marius Stephant Mar 05 '18 at 19:53
  • Examples! Do $S_{10}$, say. – Did Mar 05 '18 at 19:55
  • Hello everyone. @Did: if you do $S_{10}$ you will obviously leave some terms out, but in the above example, we have $S_n$ twice, so, isn't it the same sum? – Sal.Cognato Mar 05 '18 at 20:03
  • @Sal.Cognato The point is that "we do not have $S_n$ twice", see below for a concrete case. – Did Mar 05 '18 at 20:04

1 Answers1

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It seems you are only considering partial sums $S_n$ hence the considerations of (absolute) convergence and rearrangements are offtopic.

However, if you make the "⋯" in your post explicit, you will see that one does not reach "in essence" $S_n=\frac12S_n$. For example, $$S_{10}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10}$$ hence the reordering of terms that you suggest yields $$S_{10}=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{0}+\left(\frac17-\color{red}{0}\right)-\color{red}{0}+\left(\frac19-\color{red}{0}\right)-\color{red}{0}$$ whose value is not at all equal to $$\frac12S_{10}=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{\frac1{12}}+\qquad$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\left(\frac17-\color{red}{\frac1{14}}\right)-\color{red}{\frac1{16}}+\left(\frac19-\color{red}{\frac1{18}}\right)-\color{red}{\frac1{20}}$$ More generally, for every $n$, the reordering of the terms of $S_{2n}$ forgets $n$ terms in $\frac12S_{2n}$, which are equal to $-\color{red}{\frac1{2n+2k}}$ for $1\leqslant k\leqslant n$.

Edit: Likewise, $$S_9=1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19$$ hence the suggested reordering yields $$S_9=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\color{red}{0}\right)-\color{red}{0}+\left(\frac17-\color{red}{0}\right)-\color{red}{0}+\left(\frac19-\color{red}{0}\right)$$ whose value is not at all equal to $$\frac12S_9=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{\frac1{12}}+\qquad$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\left(\frac17-\color{red}{\frac1{14}}\right)-\color{red}{\frac1{16}}+\left(\frac19-\color{red}{\frac1{18}}\right)$$ More generally, for every $n$, the reordering of the terms of $S_{2n+1}$ forgets $n$ terms in $\frac12S_{2n+1}$, which are equal to $-\color{red}{\frac1{2n+2k+2}}$ for $1\leqslant k\leqslant n$.

Did
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  • What is k represent in your above relation? – Sal.Cognato Mar 06 '18 at 20:17
  • A running index, to describe the terms $-\frac1{2n+2}$, $-\frac1{2n+4}$, $-\frac1{2n+6}$, and so on until $-\frac1{2n+2n}$. – Did Mar 06 '18 at 20:19
  • @Did: But in his (or her) question, he refers to $S_n$, without distinguishing between odd or even n. In the general case I don't see where the (alleged) paradox lies! – Tom Galle Mar 07 '18 at 12:57
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    @TomGalle Indee there is no paradox left, once one notices that the computations leading to the identity $S_n=\frac12S_n$ are wrong, because of the missing terms mentioned in my post. Is this your point? – Did Mar 07 '18 at 16:52
  • No, my point is, why do you reorder the terms of $S_2n$ and not $S_n$? In this case, how do we depict that we leave some terms out (and therefore we are not having the same sum $S_n$)? – Tom Galle Mar 07 '18 at 16:57
  • @TomGalle ?? The reordering of $S_n$ when $n$ is odd is almost exactly the same. – Did Mar 07 '18 at 17:06
  • @Did: Can you explain why in your solution, $\frac12S_{10}$ equals to what you write after "whose value is not at all equal to"? I understand the definition of $S_{10}$ above this, but how about $\frac12S_{10}$ below (by which you deduce that some terms are missing)? Thank you! – Samuel Mar 12 '18 at 07:56
  • @Samuel If you replace by its actual value each parenthesis in the RHS of the identity you are referring to, you get the sum of $\frac1{2k}$ from $k=1$ to $k=10$, that is, $\frac12S_{10}$. – Did Mar 12 '18 at 07:59
  • Thank you. Also, "More generally, for every n, the reordering of the terms of $S_{2n+1}$ forgets n terms... etc. But why are we interested in reordering the terms of $S_{2n+1}$? My understanding is that we want to depict that in the general case of S, if we reorder the terms as in the second row, and after the calculations, we get $S=\frac12S$, which supposedly is a paradox. How do we prove it in your solution? I don't get it. Sorry for my stupid questions! – Samuel Mar 12 '18 at 08:57
  • @Samuel Because this reordering shows that $S_n\ne\frac12S_n$, due to the missing terms, and contrarily to what the question asserts. My answer also points at the reason why the question went astray, namely a confusion between manipulations on $S_n$, for fixed finite values of $n$, which are valid but do not yield $S_n=\frac12S_n$, vs. manipulations on their limit $S$, which are illicit. By the way, it seems you are falling in the exact same trap than the OP... hence I might go as far as suggesting that you reread carefully and slowly my answer. – Did Mar 12 '18 at 09:37