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We all know that the following harmonic series

$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =\frac 1 1 + \frac 12 + \frac 13 +\frac 14+ \frac 15+ \frac 16+ \cdots$$ $$> \frac 12+\frac 12+ \frac 14+ \frac 14+ \frac 16+ \frac 16+ \cdots =\frac 1 1 + \frac 12 + \frac 13 +\cdots = S.$$ In this way we see that $S > S$.

Can we conclude from this that $S$ is divergent??

User8976
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    yes, this is a valid proof (if the sum converged to a number $S$, your series manipulations would be valid, then $S > S$, which is impossible, so the assumption that the sum converged is wrong). – hunter Feb 22 '15 at 19:20
  • I'm not sure this is valid: Let $s_n$ denote the $n$ partial of the series. You are showing that $s_n > s_{n/2}$. This is something that is true for many convergent series, such as the sum of reciprocals of squares. In the limit it only gives $S \geq S$, which is not a contradiction. – Elle Najt Feb 22 '15 at 19:23
  • this is proof 6 in the following document http://stevekifowit.com/pubs/harmapa.pdf – hunter Feb 22 '15 at 19:26
  • @AreaMan, no, this argument shows that $\lim_{n \to \infty} s_n > \lim_{n \to \infty} s_{n/2}$ which is indeed a contradiction (since if the series converges, both limits are the same). – hunter Feb 22 '15 at 19:27
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    @hunter But $>$ in general becomes $\geq$ in a limit. Else $1/n > 0$ implies $0 = lim 1/n > 0$ – Elle Najt Feb 22 '15 at 19:28
  • I was wondering that such a easy proof must exist.... finally found it!! @hunter – User8976 Feb 22 '15 at 19:29
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    I agree with AreaMan's criticism of the proof as written. But note that one could insert a step to turn the inequality into the equality $S = \big( (1-\frac12) + (\frac13-\frac14) + (\frac15-\frac16) + \cdots \big) + S$, which is a clear contradiction. That equality does descend to a rigorous proof on the level of partial sums (indeed, even $s_n > (1-\frac12) + s_{n/2}$ is enough). – Greg Martin Feb 22 '15 at 19:30
  • @GregMartin How exactly can you claim that $S = \left( (1-\frac12) + (\frac13- \frac14) + (\frac15-\frac16) + \cdots \right) + S$? You assume that the harmonic series converges, OK, $S\in \mathbb R$.Then how to you rewrite $S$ as $\left( (1-\frac12) + (\frac13- \frac14) + (\frac15-\frac16) + \cdots \right) + S$? – Git Gud Feb 22 '15 at 19:38
  • \begin{align}s_{2n} &= \frac11 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{2n-1} + \frac1{2n-2} \&= \bigg(\frac12 + \frac12 + \frac14 + \frac14 + \cdots + \frac1{2n-2} + \frac1{2n-2}\bigg) \&\qquad{}+ \bigg(\Big(1-\frac12\Big) + \Big(\frac13-\frac14\Big) + \cdots + \Big(\frac1{2n-1} - \frac1{2n-2}\Big)\bigg) \&= s_{n} + \bigg(\Big(1-\frac12\Big) + \Big(\frac13-\frac14\Big) + \cdots + \Big(\frac1{2n-1} - \frac1{2n-2}\Big)\bigg);\end{align}take limits. – Greg Martin Feb 22 '15 at 19:43
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    I disagree with @Greg and AreaMan's criticism. One can prove the first inequality is strict (assuming existence of both) before one recognizes that the partial sums of the second series are in fact $s_{n/2}$, simply by comparing termwise. After that, one notices the existence of the first implies the existence of the second, and hence existence of the first implies the strict inequality. This is what I assumed OP was doing when I first read the question. – anon Feb 22 '15 at 20:04
  • @whacka But the OP didn't justify his steps. As you may have seen, I and some others interpreted that the OP was comparing partials sums instead of the actual summands. That the OP meant something else and wasn't clear about it, is the OP's fault, not the readers'. – Git Gud Feb 22 '15 at 20:19
  • Suppose that $ S $ converges. Let $ (S_{n}){n \in \mathbb{N}} $ denote the sequence of partial sums of $ S $ and $ (T{n}){n \in \mathbb{N}} $ the sequence of partial sums of $$ T = \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \cdots. $$ By the Comparison Theorem, $ T $ converges, and we also have $ T < S $. On the other hand, we have $$ \forall n \in \mathbb{N}: \quad S{n} = T_{2 n} < S_{2 n}. $$ The Squeeze Theorem then yields $ T = S $, which is a contradiction. – Berrick Caleb Fillmore Feb 22 '15 at 20:19
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    @GitGud Perhaps that's true. One could also argue about how charitable a given interpretation is and if readers have any responsibility to exert minimal effort to seek a charitable interpretation, and if that applies in this case (i.e. if finding this interpretation would have required more than low effort on readers' part). I am not sure which way to conclude. – anon Feb 22 '15 at 20:24
  • The only missing ingredient is that partial sums are strictly increasing, therefore either the series diverges to infinity or it has a limit (it cannot oscillate or have other weird behaviors). –  May 24 '19 at 18:05

5 Answers5

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The proof can be made a bit more rigorous by setting $$ \begin{align} a_n=\frac1n:&\,\quad1,\,\frac12,\frac13,\frac14,\frac15,\frac16,\dots\\b_n=\frac1{2\lfloor(n+1)/2\rfloor}:&\quad\frac12,\frac12,\frac14,\frac14,\frac16,\frac16,\dots \end{align}\tag{1} $$ Note that $a_n\ge b_n$, $a_n\gt b_n$ when $n$ is odd, and $a_n=b_{2n-1}+b_{2n}$.

Assuming that $$ \sum_{n=1}^\infty a_n\tag{2} $$ converges, then $$ \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty(b_{2n-1}+b_{2n})=\sum_{n=1}^\infty a_n\tag{3} $$ also converges. However, $$ \sum_{n=1}^\infty(a_n-b_n)\gt0\tag{4} $$ Since $a_n\ge b_n$ and $a_n\gt b_n$ when $n$ is odd.

Now, $(3)$ says that $$ \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty a_n\tag{5} $$ and $(4)$ says that $$ \sum_{n=1}^\infty b_n\lt\sum_{n=1}^\infty a_n\tag{6} $$ These last two statements are contradictory, so the assumption that $(2)$ converges must be false.

robjohn
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If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both exist, $a_n\ge b_n$ for all $n$, and $a_i>b_i$ for at least one $i$, then the first sum must be strictly greater than the second. This is because the first's partial sum is eventually always at least $a_i-b_i$ more than the second's partial sums. In this case, one can subsequently reason that if the first exists, so does the second. If this is your reasoning, it is valid.

anon
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Here is another approach to making the answer more rigorous. Assume the series converges, then $$ \begin{align} \sum_{k=1}^\infty\frac1k &=\sum_{k=1}^\infty\left(\frac1{2k-1}+\frac1{2k}\right)\\ &=\sum_{k=1}^\infty\left[\left(\frac1{2k}+\frac1{2k}\right)+\left(\frac1{2k-1}-\frac1{2k}\right)\right]\\ &=\sum_{k=1}^\infty\frac1k+\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)\\ &=\sum_{k=1}^\infty\frac1k+\log(2) \end{align} $$ The rearrangements are justified because the series are all of positive terms.

The sum $$ \begin{align} \sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\[6pt] &=\log(2) \end{align} $$ is a well-known series for $\log(2)$.

robjohn
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There's another way to approach this, via integration: compare - on the domain $(1, \infty)$ the curve $y_1={1\over x}$ with the step function $y_2={1\over Floor(x)}$ (where $Floor(x)$ is the greatest integer $<x$).

Clearly, for every $x\in (1,\infty)$, we have $0<y_1(x)\le y_2(x)$, so $\int_1^\infty y_1dx\le\int_1^\infty y_2dx$; moreover, $\int_1^\infty y_2dx$ is just the sum of the harmonic series.

But integrating, we get $\int_1^\infty y_1dx=\ln(x)\vert^\infty_1=\infty$, so the harmonic series must diverge.

Of course, this is non-rigorous, but it's good motivation, and it can be made rigorous without much work.

Noah Schweber
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The standard proof that everyone has seen but no one (or at least not i) can remember is to group more and more terms together, in such a way as to see the series is larger than $1+\frac 12+\frac 12+\dots $...

Why not include that here ?...

Namely if $S_n $ is the $n $-th partial sum, just note that $S_{2n}-S_n=\frac 1 {n+1}+\dots +\frac1{2n}\gt \frac 12 $...

Thus the sequence of partial sums is not Cauchy, so the series doesn't converge... (or by comparison with $1+\frac 12+\frac 12+\dots $)