# Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.

If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.

Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.

• It is a very strange phenomenon that many problem books seem to push the Bertrand's Postulate solution to this problem. I remember that this came up as a problem (apropos of nothing) in my freshman year math class, and I had some problem book at hand and duly turned in a solution which used BP. The next year I got the problem in a number theory course and by then was sophisticated enough to see the elementary solution involving the ord_2 function. Commented Aug 19, 2010 at 0:04
• Note that I include this exercise as a -- not fully worked out -- example in my (relatively advanced) undergraduate number theory course. See the example on page 13 of math.uga.edu/~pete/4400intro.pdf. (I should admit that a lot of the students have trouble with the corresponding homework problem that asks the details to be filled in.) Commented Aug 19, 2010 at 0:05
• @Pete: that's interesting. In high school competition math circles the 2-adic proof is very well known. I first learned it on the AoPS website but it is probably also in some competition book. Commented Aug 19, 2010 at 2:07
• I remember that in my first semester I was asked about it and looking in some books I always arrived to the Bertrand postulate. But if you think so, Bertrand Postulate is still harder to prove. Commented Jul 29, 2015 at 1:44
• This video might help.
– user840532
Commented Oct 28, 2020 at 9:27

Hint  There is a $$\rm\color{darkorange}{unique}$$ denominator $$\rm\,\color{#0a0} {2^K}$$ having maximal power of $$\:\!2,\,$$ so scaling by $$\rm\,\color{#c00}{2^{K-1}}$$ we deduce a contradiction $$\large \rm\, \frac{1}2 = \frac{c}d \,$$ with odd $$\rm\,d \:$$ (vs. $$\,\rm d = 2c),\,$$ e.g.

$$\begin{eqnarray} & &\rm\ \ \ \ \color{0a0}{m} &=&\ \ 1 &+& \frac{1}{2} &+& \frac{1}{3} &+&\, \color{#0a0}{\frac{1}{4}} &+& \frac{1}{5} &+& \frac{1}{6} &+& \frac{1}{7} \\ &\Rightarrow\ &\rm\ \ \color{#c00}{2}\:\!m &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &+&\, \color{#0a0}{\frac{1}{2}} &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}\\ &\Rightarrow\ & -\color{#0a0}{\frac{1}{2}}\ \ &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &-&\rm \color{#c00}{2}\:\!m &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M} \end{eqnarray}$$

All denom's in the prior fractions are odd so they sum to fraction with odd denom $$\rm\,d\, |\, 3\cdot 5\cdot 7$$.

Note  Said $$\rm\color{darkorange}{uniqueness}$$ has easy proof: if $$\rm\:j\:\! 2^K$$ is in the interval $$\rm\,[1,n]\,$$ then so too is $$\,\rm \color{#0a0}{2^K}\! \le\, j\:\!2^K.\,$$ But if $$\,\rm j\ge 2\,$$ then the interval contains $$\rm\,2^{K+1}\!= 2\cdot\! 2^K\! \le j\:\!2^K,\,$$ contra maximality of $$\,\rm K$$.

The argument is more naturally expressed using valuation theory, but I purposely avoided that because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student.

See the Remark here for a trickier application of the same idea (from a contest problem).

• did you try to show for $n=7$? Commented Jul 31, 2013 at 4:22
• I am not able to understand how to use your hint for general $n$ Commented Jul 31, 2013 at 4:26
• Beautiful! Much nicer than an ugly argument in the style of "writing everything as one fraction, one 'sees' that the denominator has more factors $2$" but essentially the same, though. Commented Jun 21, 2014 at 15:17
• Please elaborate for n=8,16 etc Commented May 13, 2020 at 8:46
• you also have to show that no other denominator contains $2^k$ as a factor. Commented Oct 20, 2021 at 22:47

An elementary proof uses the following fact:

If $2^s$ is the highest power of $2$ in the set $S = \{1,2,...,n\}$, then $2^s$ is not a divisor of any other integer in $S$.

To use that,

consider the highest power of $2$ which divides $n!$. Say that is $t$.

Now the number can be rewritten as

$\displaystyle \frac{\sum \limits_{k=1}^{n}{\frac{n!}{k}}}{n!}$

The highest power of $2$ which divides the denominator is $t$.

Now the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is at least $t-s$. If $k \neq 2^{s}$, then this is atleast $t-s+1$ as the highest power of $2$ that divides $k$ is atmost $s-1$.

In case $k=2^s$, the highest power of $2$ that divides $\dfrac{n!}{k}$ is exactly $t-s$.

Thus the highest power of $2$ that divides the numerator is atmost $t-s$. If $s \gt 0$ (which is true if $n \gt 1$), we are done.

In fact the above proof shows that the number is of the form $\frac{\text{odd}}{\text{even}}$.

• It would probably be a good idea to flesh this out a little. Commented Aug 18, 2010 at 22:58
• The exact same proof I gave works, just use $2^k$ instead of $p$. Again, you get that the sum is of the form $\frac{1}{2^k}+\frac{a}{b}$, where $b$ (being a divisor of the lcm of stuff divisible by at most $k-1$ copies of 2) is not divisible by $2^k$. This can't be an integer, otherwise $\frac{b}{2^k}+a$ would be an integer, which it isn't. Commented Aug 18, 2010 at 23:00
• But what is there to say the sum of the $n!/k$s do not have $2^k$ dividing them? Commented Jan 28, 2021 at 7:22

I never heard of the Bertrand postulate approach before, although it turns out that the first proof that the $$n$$-th harmonic sum is not an integer when $$n > 1$$ uses Bertrand's postulate and determinants. It appeared in a paper of Theisinger (Bemerkung über die harmonische Reihe, Monatsh. f. Mathematik und Physik 26 (1915), 132--134) that you can read here and he refers to Bertrand's postulate as Chebyshev's theorem. (Update: in several places I have seen Theisinger misspelled as Taesinger, and I am guilty of doing that myself in this answer until I corrected it.) The 2-adic proof is due to Kürschák (A Harmonikus Sorról, Mat. és Fiz. Lapok, 27 (1918), 299--300) and you read it here.

I like to think of this result as saying the $$n$$-th harmonic sum tends to infinity $$2$$-adically. That naturally raises the question of the $$p$$-adic behavior of harmonic sums for odd primes $$p$$, which quickly leads into unsolved problems. I wrote a discussion of that here.

• +1: a new blurb. If there were a listserve that would automatically notify me whenever there is a new blurb from KCd, I would gladly sign up for it! Commented Aug 19, 2010 at 3:56
• @PeteL.Clark : You can create a feed for a webpage with pagee2rrss.com. In this case, add to you feeds reader: page2rss.com/rss/2b74436a91d372fca18b5f1645f1d59e Commented Dec 5, 2011 at 19:58
• @leonbloy: Thannks, that sounds useful. I'll give it a try... Commented Dec 6, 2011 at 4:01

What the heck -- I'll leave my comment as an answer.

See the Example on p. 13

This is discussed, together with (as a footnote) the strange phenomenon that this is often solved by an appeal to Bertrand's Postulate. The discussion in the above text is intended to be "didactic" in that a few details are left to the reader, and I recommend it as a good exercise to flesh them out.

• @Pete: But your exposition uses valuation theory - which disqualifies it as "elementary". Obviously the problem is trivial to anyone knowing valuation theory.Namely the sum has a lone dominant term with $v_2 < 0$ so, by the domination principle, the sum has $v_2 < 0$ so is nonintegral. Commented Aug 19, 2010 at 0:36
• @BD: The construction uses something that happens to be a valuation, but I don't think that makes it valuation theory. The definition of ord-p uses nothing more or less than the fundamental theorem of arithmetic, so is appropriate in an introductory course. The point of this exercise is to get students used to making arguments of this kind which -- if they continue on in their study of number theory -- will be seen to be valuation-theoretic. (Anyway the argument can certainly be phrased without using ord-2 if that's your taste.) Commented Aug 19, 2010 at 2:49
• By the way, the fact that each partial sum has a unique term of minimal 2-adic valuation was not so easy for my students. That's most of the exercise that I left for them to solve, and for many it was challenging, not trivial. Commented Aug 19, 2010 at 2:53
• @Pete: Certainly it can be presented at an elementary level, and there is no doubt that it is instructive to do so. However, it's a lot of overhead to introduce for a single problem - as here. It's puzzling to hear that the exercise proved difficult for some students. Did you give them prior examples of employing the dominance principle? E.g. that a set of integers having precisely one odd element has odd sum (that is precisely what occurs in the sum above if one multiplies it through by a 2-least common denominator). Commented Aug 19, 2010 at 3:17
• Hi BS, I understand your point. But I think, in this kind of forum, this type of answer have a great value. Altought Anton asked for simple proof, the Pete's argument help the interested students to learn most power techniques in simple situation. @Thanks Pete for the link. Commented Aug 19, 2010 at 3:25

This is a h.w. problem in Ch 1 of "Ireland and Rosen" - prob 30. There is a hint on p. 367. Let $s$ be the largest integer such that $2^s \le n$, and consider:

$\sum \limits_{k=1}^n \frac{2^{s - 1}}k$

Show that this sum can be written in the form $a/b$ + $1/2$ with $b$ odd.

Then apply problem 29 which is:

Suppose $a, b, c, d$ in $\mathbb{Z}$ and $gcd (a,b) = (c,d) = 1$

If $(a/b) + (c/d)$ = an integer, then $b = \pm d$. (But $b$ odd, $d$ = $2$.)

Maybe this was part and parcel of earlier answers. If so forgive me for trying.

I kind of have an elementary solution, it seems to be fine but I am not sure if everything is correct; please point out the mistake(s) I'm making, if any.

Define $$H_n:=\sum_{i=1}^n \frac{1}{i}$$ Since $0<H_n<n$, if $\exists$ some $n$ for which $H_n$ is integral then $H_n=k$ where $0<k<n$. Then $$H_n=k=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{k}+\cdots\ +\frac{1}{n}\\ \Rightarrow k=\frac{1}{k}+\frac{p}{q}\Rightarrow qk^2-pk-q=0$$ where $\gcd(p,q)=1$. Then we get $$k=\frac{p\pm \sqrt{p^2+4q^2}}{2q}$$ Since $k$ is integer $$p^2+4q^2=r^2$$ for some $r\in \mathbb{Z}^+$. Let $\gcd(p,2q,r)=d$ and let $\displaystyle x=\frac{p}{d},\ y=\frac{2q}{d},\ z=\frac{r}{d}$. Then $$x^2+y^2=z^2$$ Now, I make the following claim:

Claim:$p$ is odd and $q$ is even.

Proof: Let $s=2^m\le n$ be the largest power of $2$ in $\{1,2,\cdots,\ n\}$. Then, if $k\ne s$ then the numerator of $\displaystyle \frac{p}{q}$ is the sum of $n-1$ terms out of which one will be odd and hence $p$ is odd. On the other hand, $q$ will have the term $s$ as a factor. So q is even.

Now, if $k=s$, then since $n>2$(otherwise there is nothing to prove)then, there will be a factor $2^{m-1}\ge 2$ in $q$ and one of the sum terms in $p$ that corresponds to $2^{m-1}$ will be odd. Hence in this case also, $p$ is odd and $q$ is even. So the claim is proved. $\Box$

So, now we see that $d\ne 2$ and hence $2|y$. So we have a Pythgorian equation with $2|y, \ x,y,z>0$. hence the solutions will be $$x=u^2-v^2,\ y=2uv,\ z=u^2+v^2$$ with $(u,v)=1.$ So, since $k$ is positive, $$k=\frac{d(x+z)}{dy}=\frac{u}{v}$$ But since $(u,v)=1$, $k$ is not an integer (for $n\ge 2$) which is a contradiction. So $H_n$ can not be an integer. $\Box$

• I know this is a very late answer, but in the proof of your claim you said that p is the sum of n-1 terms out of which one is odd. Which one is that? And was it really intentional to write the requirement (p,q)=1? Commented Aug 17, 2018 at 0:01

Very similar to the Bertrand approach, except significantly more elementary.

Suppose for contradiction that a partial sum of the harmonic series is an integer $z$:

$$1 + \frac{1}{2} + \frac{1}{3}+...+\frac{1}{n}=z$$

Now consider the maximal power of $2$ below $n$ and let's call it $2^t$. (Note that all other integers between 1 and $n$ have a power of $2$ strictly less than t). Now consider the unique prime factorization of $n!$. The exponent of $2$ in this factorization will be greater than or equal to $2^t$, but instead let us define $M$ as $n!$, except with the power of $2$ in its prime factorization set to be $t-1$ (as opposed to some integer greater than $t$).

Multiply both sides of the equation by $M$:

$$M+\frac{M}{2}+\frac{M}{3}+...+\frac{M}{2^t}+...+\frac{M}{n}=Mz$$

$M$ has enough factors to make all terms on the LHS integers except for the $\frac{M}{2^t}$ term. Summing the LHS, we see that is not an integer, even though the RHS is an integer. Contradiction, QED.

This proof is essentially the same as the proof with Bertrand's postulate, except with $2^t$ instead of a prime number $p$ between $\frac{n}{2}$ and $n$.

• This same approach was already described in several other answers. Commented Jul 29, 2015 at 1:31

Here's a short proof: Let $H_n = \displaystyle \sum_{k=1}^n\dfrac{1}{k}.$ One can show that $\displaystyle\sum_{k=1}^{n}\dfrac{(-1)^{k-1}\binom{n}{k}}{k}= H_n.$ This can be rewritten as: $$\sum_{k=0}^{n}{(-1)^k\binom{n}{k}a_k} = b_n$$

where $a_0 =0$ and $a_i = \dfrac{1}{i}$ for $i=1,\ldots n$ and $b_n = -H_n$

This answer shows that the $b_i$ are integers if and only if the $a_i$ are integers. Clearly for $i \geq 2$ we can see that the $a_i$ are not integers, from which it follows that neither are the $b_i, i\geq 2.$

I feel like the answers so far still aren't as basic as the one sketched by Mathologer in https://youtu.be/vQE6-PLcGwU?si=0nNu0EF6wkvGEeOA&t=1046. It is the same claim as some of the previous answers, but the following proof I feel is a little conceptually simpler, or written in more basic language (not as clever/"magical" as Bill Dubuque's highly upvoted answer, but hopefully something more down to earth, like something an intro-to-proofs student could write just after noticing the basic pattern that the partial sums are all of the form $$\frac{[\text{odd}]}{[\text{even}]}$$. Also has the added benefit of being a direct proof a stronger fact, instead of a proof by contradiction.):

Claim: writing as a fraction in lowest terms, the $$n$$th partial sum $$S_n := 1+\frac 12 + \ldots + \frac 1n$$ is of the form $$\frac{[\text{odd}]}{[\text{even}]}$$ ($$n>1$$).

Proof: by strong induction. Base case: $$n=2$$, $$S_2=\frac 32$$. By the inductive hypothesis, for $$n>2$$, we now assume $$S_{n-1}$$ is of the form $$\frac{[\text{odd}]}{[\text{even}]}$$.

• Case 1: $$n$$ is odd, in which case $$S_n= S_{n-1} + \frac 1n = \frac{[\text{odd}]}{[\text{even}]} + \frac 1n = \frac{([\text{odd}] \cdot n) + (1\cdot [\text{even}])}{[\text{even}] \cdot n} = \frac{[\text{odd}]}{[\text{even}]},$$ (using that odd times odd is odd, plus even is still odd) which remains true even after reducing the fraction to lowest terms.
• Case 2: $$n$$ is even, in which case we can split $$S_n$$ into a sum over even and odd numbers, $$S_n = E_n + O_n$$ where $$E_n := \frac 12 + \frac 14 + \ldots + \frac 1n$$, and $$O_n = S_n - E_n = \frac 11 + \frac 13 + \ldots + \frac 1{n-1}$$. Then $$E_n = \frac 12 \cdot S_{n/2}$$, which by our induction hypothesis, is of the form $$\frac{[\text{odd}]}{[\text{even}]}$$. And we know by Case 1 that adding things of the form $$\frac{1}{[\text{odd}]}$$ to things of the form $$\frac{[\text{odd}]}{[\text{even}]}$$ preserve that form, so indeed $$S_n = E_n + O_n$$ is of the form $$\frac{[\text{odd}]}{[\text{even}]}$$.

This proof by strong induction can also prove that any difference of partial sums, $$\sum_{k=n_1}^{n_2} \frac 1k$$ is also of the form $$\frac{[\text{odd}]}{[\text{even}]}$$, and hence is not an integer either. (Bill Dubuque's slick 3 line proof can also prove this; it can also show the Claim in my answer, since using his notation, he shows that $$2^K m = \frac 12 + \sum \frac{[\text{even}]}{[\text{odd}]}$$, and $$\frac 12 = \frac{[\text{odd}]}{[\text{even}]}$$ plus any $$\frac{[\text{even}]}{[\text{odd}]}$$ will result in $$\frac{[\text{odd}]}{[\text{even}]}$$. Dividing by $$2^K$$ again keeps it as $$m=\frac{[\text{odd}]}{[\text{even}]}$$.)

• Bill Dubuque's answer looks basic enought to me, anyway also the one you are proposing is very fine!
– user
Commented Jan 21 at 20:09
• @user yes, but I feel perhaps it is a bit too clever, “too slick”. Phrased as a proof by strong induction, it may feel more familiar to students just starting to learn about proofs (especially since we teach students that when asked to prove something is true for all $n$, induction is the first thing to try)
– D.R.
Commented Jan 21 at 20:46

A more general approach that includes the proof using the prime 2 but is valid for any prime $<n$ (posted elsewhere with an erroneous n! instead of LCD): Let the least common denominator of the harmonic series H(n) be LCD(n). Take any prime p in the sequence 1 to n and let q be the highest power of p so that $p^q ≤ n$.

For any k, $1 ≤k ≤n$, LCD(n)/k is an integer and = 0 (mod p) except $LCD(n)/p^q$ which is an integer and does not contain p, and therefore cannot be 0 (mod p). But H(n)LCD(n)=0 (mod p) (since LCD(n) contains the factor p), a contradiction if H(n) is an integer.

(The simplicity comes from the use of a complicated LCD(n) which exists but whose prime powers I would not be able to describe in the general case).

if we consider highest prime upto $n$ then given sum can be written as $1/p + a/b$ where a is some integer $b$ is also an integer not divisible by $p$. so $b/p$ can not be an integer and so $b/p + a$. so the given sum can not an integer

• For that conclusion, you need to know that $2p > n$, that is, you need Bertrand's postulate. Commented Oct 15, 2015 at 11:39
• You might want to use MathJax. The simple way to use it is to surround your math with \\$. There are some commands you might need to learn. Commented Oct 15, 2015 at 11:46

If $$n \geq 2$$, let $$f = x + \frac{1}{2}x^2 + \cdots + \frac{1}{n}x^n$$ in $$\mathbb Z[x]$$. Suppose that there exists a $$k \in \mathbb Z$$ such that $$k - f(1) = 0$$. Then $$(x-1)$$ divides $$k-f$$ so we may write $$k-f = (x-1)g$$ where $$g \in \mathbb Z[x]$$ and $$\deg g = n-1$$.

Taking the formal derivative of both sides, we obtain $$-f' = (x-1)g' + g$$. If $$a$$ is the leading coefficient of $$g$$, then since $$\deg g = n-1$$ we have $$a + (n-1)a = -1$$ since $$f' = 1 + \cdots + x^{n-1}$$. In particular, we have $$an = -1$$, which is impossible since $$n \geq 2$$ and $$a$$ is an integer.