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Bartle & Sherbert, edition 4, page-97 gives the following proof:

Assume that the series converges to some $S$:

$$\implies S= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \dots$$

Then they proceed to derive an inequality by converting the terms of the type $\frac{1}{2n-1}+\frac{1}{2n}$ into $\frac{1}{2n}+\frac{1}{2n}=\frac{1}{n}$ $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \dots > \left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right) +\dots=1+\frac{1}{2}+\frac{1}{3} \dots=S$$ They have shown $S>S$ from the assumption, hence, there's a contradiction.

But I do not like their insistence upon using the "$\dots$", as I feel like they're hiding something behind that.

If we do a similar analysis a bit more rigorously, i.e, working solely with sequences of partial sums up to a defined, fixed $n$, (instead of those dots), we have:

$$S_{2n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \dots +\frac{1}{2n-1}+\frac{1}{2n}$$ $$S_{2n} >\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right) \dots +\left(\frac{1}{2n}+\frac{1}{2n}\right)=1+\frac{1}{2}+\frac{1}{3} \dots \frac{1}{n}=S_n \implies S_{2n} > S_{n}$$ This is hardly news, since the 1-harmonic is monotone increasing.

Now assuming that the sequence $S_n$ converges, and passing onto the limit in the above equation yields: $$S \geq S$$ which is fine?

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    $\begingroup$ @H.sapiensrex That's not true. Many examples of conditional convergence series leading to contradictions when not handled properly comes from that way of using ellipsis. $\endgroup$
    – jjagmath
    Commented Jan 21 at 11:20
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    $\begingroup$ @ChrisMzz. Because when dealing with series you can't just group the terms like that and expect the resulting series to have the same sum. $\endgroup$
    – jjagmath
    Commented Jan 21 at 11:30
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    $\begingroup$ In fact, if a series is convergent but not absolutely convergent then for any real x you can order the series so it converges to x (!) $\endgroup$
    – David Raveh
    Commented Jan 21 at 11:33
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    $\begingroup$ @ChrisMzz. The problems with series arises not only when permuting the terms, but also with grouping them in different ways (introducing parenthesis). $\endgroup$
    – jjagmath
    Commented Jan 21 at 11:44
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    $\begingroup$ Here S is assumed to be a finite number. If the terms of the series are positive, absolute convergence is the same as convergence. Permuting or grouping should not matter when there is absolute convergence. And this is talking about a hypothetical S itself, which is not "$S_n$" or "$S_{2n}$" $\endgroup$
    – qrsngky
    Commented Jan 22 at 2:20

6 Answers 6

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The estimate $S_{2n} > S_{n}$ is indeed not good enough to prove that the harmonic series is not convergent. But with a small modification one can show that $S_{2n} > a + S_{n}$ for some positive number $a$ which does not depend on $n$, and that suffices for the proof.

Concretely: For all $n\ge 2$ is $$ \begin{align} S_{2n}&=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \dots +\frac{1}{2n-1}+\frac{1}{2n} \\ &>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right) \dots +\left(\frac{1}{2n}+\frac{1}{2n}\right) \\ &= \frac 12 + S_n \, . \end{align} $$ If the harmonic series were convergent with value $S$ then both $S_{2n}$ and $S_n$ would converge to $S$, and we could conclude that $$ S \ge \frac 12 + S $$ which is of course impossible.

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    $\begingroup$ The endgame can be simplified a bit: Once you’ve shown that $S_{2n} > \frac{1}{2}+S_n$, you can immediately conclude $S_{2^k} \geq \frac{k}{2}$, giving a direct proof of divergence with a concrete lower bound on the rate. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jan 22 at 19:56
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    $\begingroup$ @PeterLeFanuLumsdaine: Yes, I am aware of that. The question (as I understand it) was how the argument “$S_{2n} > S_n \implies S \ge S \implies \sum 1/n \text{ is divergent}$” can be made rigorous, and that is what I tried to answer. – Of course there are many other proofs for the divergence of the harmonic series. $\endgroup$
    – Martin R
    Commented Jan 22 at 20:01
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    $\begingroup$ Re, of course it was just a typo, but your $S \ge S$ should be $S > S$. $\endgroup$
    – LSpice
    Commented Jan 22 at 23:18
  • $\begingroup$ @LSpice Well, the original issue was that the OP interpreted Bartle & Sherbert's argument as $S_{2n}>S_n \Rightarrow S \geq S$, which does not imply that $\sum\frac{1}{n}$ is divergent. That's because $S_{2n}>S_n \Rightarrow \lim_{n\rightarrow\infty}S_{2n} \geq \lim_{n\rightarrow\infty}S_n$ (here $>$ became $\geq$ after taking the limit). However, what was needed for a contradiction was really $\lim_{n\rightarrow\infty}S_{2n} > \lim_{n\rightarrow\infty}S_n$ (with strict $>$, so they have finite difference). Bartle & Sherbert did not claim that "..." were finitely many terms, though. $\endgroup$
    – qrsngky
    Commented Jan 25 at 1:55
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Another explanation leading to $S_{2n}>S_n+{1\over 2}$ could be $$S_{2n}=\left (1+{1\over 2}+\ldots +{1\over n}\right )+\left ({1\over n+1}+\ldots +{1\over 2n} \right )\\ > S_n +\underbrace{{1\over 2n}+\ldots +{1\over 2n}}_{n\ {\rm terms}}=S_n+{1\over 2}$$ This method is usually applied in proofs of the condensation test.

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You are correct. The manipulation of series like that needs some more details to be complete. In this case one needs the following theorem: if a series is absolutely convergent, then the terms of the series may be grouped (and permuted) and the sum will be the same. Since this series has only positive terms, assuming that it is convergent implies the absolute convergence and the theorem can be applied to get the contradiction.

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  • $\begingroup$ It's probably more appropriate for me to comment here to ask this, the Riemann series theorem only mentions permuation, and the inequality is built off of terms in the same order here. $\endgroup$
    – Chris Mzz.
    Commented Jan 21 at 11:40
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    $\begingroup$ @ChrisMzz. The problems with series arises not only when permuting the terms, but also with grouping them in different ways (introducing parenthesis). $\endgroup$
    – jjagmath
    Commented Jan 21 at 11:45
  • $\begingroup$ @jjagmath, how so? If we are talking about the same thing here, then the sequence of partial sums of the series with additional parentheses is a subsequence of the sequence of partial sums of the original series. $\endgroup$
    – Carsten S
    Commented Jan 22 at 18:13
  • $\begingroup$ @CarstenS Yes, we are dealing with the sequence of partial sums. And it can happen that a sequence $s_n$ diverge while one of its subsequence $s_{n_k}$ converge. And even can happen that two subsequences $s_{n_k}$ and $s_{m_k}$ converge, but to different limits. Those facts are not really important here since in this case we are beggining with a convergent sequence, and Theorem: If a sequence converge, all of its subsequences must converge to the same limit. But that needs to be proved!, and that is hidden by the use of ellipsis. $\endgroup$
    – jjagmath
    Commented Jan 22 at 19:01
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    $\begingroup$ Adding to my point: We already agreed that when a series is convergent then one can introduce parenthesis (group the terms of the series) and the resulting series will converge to the same limit. But then someone who is new to the subject, could think that something like changing $1+0+0+0+0 +\cdots$ can be replaced by $1 + (0) + (0) + (0) + \cdots$ (you are just introducing parenthesis) and then, since obviously $0 = 1-1$, you can replace that with $1 + (1-1) + (1-1) + (1-1)+ \cdots$ and now we are in a gray area. Can we drop the parenthesis? We begin with a convergent sequence after all. $\endgroup$
    – jjagmath
    Commented Jan 22 at 19:27
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I do not like his insistence upon using the "…", as I feel like he's hiding something behind that.

Of course they are hiding something behind that. Mainly, it is formal notation for the sum of a series. But the point is not to deceive, and the conclusion is not wrong. The point is exactly to present the idea of the proof in a simple, easily digestible form, without going into all the details. One can make mistakes by skipping a rigorous proof in such a case, but in context, it is reasonable to assume that the authors are summarizing a well known proof, as opposed to trying to be unreasonably loose.

The authors' proof sketch could indeed be filled in a bit to provide more rigor. That might go something like this:

Suppose that $S$ is a real number such that $S = \sum\limits_{i = 1}^\infty{1 \over i}$.

The sum of an infinite series is the limit of its partial sums, so $S = \lim_{n \to \infty}S_n$, where $S_n = \sum\limits_{i = 1}^n{1 \over i}$.

Now consider those $S_n$ where $n$ is even. These finite sums can be regrouped to express $S_n$ in the form $S_n = \sum\limits_{i = 1}^{n / 2}\left({{1 \over {2i - 1}} + {1 \over {2i}}}\right)$.

If it exists (as we have assumed), the limit of all the partial sums must also be the limit of this subset of the partial sums, so we are justified in writing $S = \sum\limits_{i = 1}^{\infty}\left({{1 \over {2i - 1}} + {1 \over {2i}}}\right)$.

But observe that for every integer $i > 0$, we have that ${{1 \over {2i - 1}} > {1 \over {2i}}}$, and therefore that $\left({{1 \over {2i - 1}} + {1 \over {2i}}}\right) > \left({{1 \over {2i}} + {1 \over {2i}}}\right) = {1 \over i}$.

With every term of $S = \sum\limits_{i = 1}^{\infty}\left({{1 \over {2i - 1}} + {1 \over {2i}}}\right)$ being strictly greater than ${1 \over i}$, it follows that $S > \sum\limits_{i = 1}^{\infty}{1 \over i} = S$.

But no real number is greater than itself, so there can be no real number $S$ satisfying $S = \sum\limits_{i = 1}^\infty{1 \over i}$. That is, the sum does not converge.

You might see steps in that proof that you would like to see proven themselves. That would be fair, and you are welcome to proceed, to whatever level of detail you like.

As for your own analysis, it is neither wrong nor inconsistent with the above, but it proves a much weaker result: that if the sum of the harmonic series is a well defined real number, then that number is equal to itself (since it cannot be strictly greater). Of course, we don't need to play with partial sums to come to that conclusion.

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    $\begingroup$ This is wrong. You discarded too much information. $a_n=\frac1n$ and $b_n=\frac1{2n}$ converge both to the same limit zero, but $a_n>b_n$. What is possible to say is that $$\frac1{2i-1}+\frac1{2i}=\frac1{i}+\frac1{2i(2i-1)},$$ and the second term presents a real difference at $i=1$ that becomes only larger in the accumulation of the later terms. $\endgroup$
    – Lutz Lehmann
    Commented Jan 23 at 17:53
  • $\begingroup$ @LutzLehmann, (i) this is a formalization of Bartle & Sherbert's argument referenced by the OP, which conceivably could be wrong. However, (ii) I think the argument presented here says just what you said, in less detail. That is, that there are differences $> 0$ between corresponding terms in the two summation formulae, always favoring the second formula over the first. $\endgroup$
    – John Bollinger
    Commented Jan 23 at 18:49
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    $\begingroup$ "That is, that there are differences >0 between corresponding terms in the two summation formulae" The point is that it's not sufficient to say $a_n-b_n$ is always a positive number, you have to show that there is a positive number that it's always greater than. $\endgroup$
    – Acccumulation
    Commented Jan 24 at 4:19
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    $\begingroup$ @Acccumulation, is there perhaps some confusion here between the terms of the sums and their series of partial sums? Consider two sums of the form $A=0+t_1+t_2+...$ and $B=1+t_1+t_2+...$. Here, $b_n−a_n$ is greater than 0 for only one pair of corresponding terms, but if these sums converge then surely $B=A+1>A$. The difference between corresponding partial sums must always equal or exceed some positive constant for $B>A$, but that does not require the same of the difference between corresponding terms. $\endgroup$
    – John Bollinger
    Commented Jan 24 at 5:31
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    $\begingroup$ Well, "$S_{2n}>S_{n}$" is not enough to produce a contradiction, which is also shown in the OP. But I don't think that's what Bartle & Sherbert tried to argue. Also, if $(a_k-b_k)>0$ for every term then that means $(a_1-b_1)>0$, $(a_2-b_2)>0$ and so on. Then $\sum_{k=1}^{n} (a_k-b_k)$ for $k>1$ is always greater than $(a_1-b_1)$ which is positive. $\endgroup$
    – qrsngky
    Commented Jan 24 at 10:02
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Suppose $\exists L: \forall \epsilon \exists N: (n>N) \rightarrow (|L-\sum_{i=0}^n a_n|<\epsilon)$

Pick $\epsilon = \frac 18$, and let $n>N$, where $N$ is an integer satisfying the requisite property above. Then $2n$ will also be greater than $N$, so $|L-\sum_{i=0}^{2n} a_n|<\epsilon$. Applying the Triangle Inequality, $\sum_{i=0}^{2n} a_n<L+\frac 18$ and $\sum_{i=0}^n a_n>L-\frac 18$.

But by the argument you cite in your question, $\sum_{i=0}^{2n} a_n> \frac 12+\sum_{i=0}^n a_n>\frac 12+L-\frac18=L+\frac38$. Thus we have $\sum_{i=0}^{2n} a_n>L+\frac38$ but also $\sum_{i=0}^{2n} a_n<L+\frac18$, wihch is a contradiction.

In summary, the definition of convergence constrains the behavior of the sum not only once we reach $N$, but also for every $n$ after $N$, including $2N$. So we can take arbitrarily large number of terms, and the argument you cite is not invalid simply because it needs a larger number of terms in one part than it has in another.

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Proposition: The harmonic series diverges.

Proof: Suppose the series converges. Let S be the series sum.

Rewrite as sums of terms with even denominator and odd denominator: S = S_odd + S_even

Note first that S_even = (1/2) * S. Second, 1/(2n-1) > 1/(2n), so S_odd > S_even.

Thus, we have the contradiction that S = S_odd + S_even > (1/2)S + (1/2)S = S.

I had that proof published over a quarter-century ago. See:

Michael W. Ecker, Divergence Of The Harmonic Series By Rearrangement, The College Mathematics Journal, May 1997, Vol. 28, No. 3, p. 209-210.

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