# Alternating harmonic series convergence

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?

• To arrange terms, it must be absolutely convergent. Mar 5, 2018 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. Mar 5, 2018 at 19:47
• 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 5, 2018 at 19:47
• @Did: I thought so, but which are the terms that I've left out? Mar 5, 2018 at 19:53
• Examples! Do $S_{10}$, say.
– Did
Mar 5, 2018 at 19:55

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 $1\leqslant k\leqslant n$.

• What is k represent in your above relation? Mar 6, 2018 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 6, 2018 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! Mar 7, 2018 at 12:57
• @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 7, 2018 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 $S_n$)? Mar 7, 2018 at 16:57