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

Question

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.

Answer 1

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.

Generally we can similarly prove that a sum of fractions is nonintegral if the highest power of a prime $\,p\,$ in any denominator occurs in $\rm\color{darkorange}{exactly\ one}$ denominator, e.g. see the Remark here where I explain how it occurs in a trickier multiplicative form (from a contest problem). In valuation theory, this is a special case of a basic result on the valuation of a sum (sometimes called the "dominance lemma" or similar). Another common application occurs when the sum of fractions arises from the evaluation of a polynomial, e.g. see here and its comment.

Answer 2

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 we consider the numerator, as we iterate through the values of $k$ from $1$ to $n$. If $k \neq 2^{s}$, then the highest power of 2 that divides $\dfrac{n!}k$ is at least $t-(s-1)=t-s+1$, as the highest power of $2$ that divides $k$ is at most $s-1$.

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 at most $t-s$. 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 $\frac{\text{odd}}{\text{even}}$.

Answer 3

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.

Answer 4

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.

Answer 5

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. $\Box$

Answer 6

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.

Answer 7

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 $n$.

Answer 8

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 $\frac{[\text{odd}]}{[\text{even}]}$.

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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 $m=\frac{[\text{odd}]}{[\text{even}]}$.)

Answer 9

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 $b_i, i\geq 2.$

Answer 10

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).

Answer 11

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

Answer 12

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.

Answer 13

Assume the $n$-th harmonic sum is an integer.

Suppose $2^s$ is the maximum power of $2$ less than $n$.

Let $Q$ be the set of all positive integers that aren't power of two not greater than $n$.

$gcd\ Q$ must have $2$-adic valuation of less than $s$, since no non-power of two less than $n$ is divisible by $2^s$.

$\sum 1/n - \sum_{q \in Q} 1/q$ is $1-1/2^s$. But that would mean that $p$-adic valuation of $gcd\ Q $ is $s$ if the harmonice sum is an integer. CONTRADICTION!!

Answer 14

Actually there is a very elegant proof that doesn't concern much advanced knowledge . Let $k$ be the largest natural number such that $2^k\leqslant n$, and $P$ be the product of all odd numbers less than or equal to $n$.In this case $$2^{k-1}PS_n=2^{k-1}P(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n})$$ , where all terms except $2^{k-1}P2^{-k}$ are integers.(It's an exercise that I found while reading a Japanese introductory number theory book.)

Answer 15

Here is my attempt at an elementary solution by induction, assuming that one who sees this in a first chapter of a first course on number theory didn't notice the cleaner and as elementary proof by strong induction given above. I guess this is more a naive "nuts and bolts" proof.

0. Let $S(n)$ be the $n$-th partial sum of the harmonic series.

1. As a base case, note that $S(2)=3/2$ isn't an integer, and assume that $S(n-1)$ isn't an integer for some $n > 2$.

2. Write $S(n-1) = a/b$, where $a$ and $b$ are coprime, and $b \neq 1$.

3. Then $S(n) = \frac{an+b}{bn}$ which is in $\mathbb{Z}$ only if $b$ and $n$ divide $an+b$.

4. But $b \mid an+b$ implies that $b \mid an$ which, in turn, implies that $b \mid n$ by Euclid's generalized lemma. Similarly, $n \mid b$, whence $b = n$.

5. Now recursively reconstruct the partial sums: \begin{align} S(n-2) &= \frac{a}{n}-\frac{1}{n-1} = \frac{a(n-1)-n}{n(n-1)} \\ \phantom{M} \\ S(n-3) &= \frac{a(n-1)(n-2)-n[(n-1)+(n-2)]}{n(n-1)(n-2)} \\ &\ \vdots \\ S(n-k) &= \frac{a\Pi_{i=1}^{k-1} (n-i) - n\sum_{i_1 < \cdots < i_{k-2}} (n-i_1)\cdots(n-i_{k-2})}{n(n-1)\cdots(n-k)} \end{align}

So then we have \begin{align} 1 = S(1) &= \frac{a(n-1)! - n \sum_{i_1 < \cdots < i_{n-2}}(n-i_1)\cdots(n-i_{n-2})}{n!} \\ &= \frac{a}{n} - [(n-1)+(n-2)+ \cdots + 2] \\ &= \frac{a}{n} - \frac{n(n-1)}{2}+1 \end{align}

Therefore, $\frac{a}{n} = \frac{n(n-1)}{2}$ which implies that $2a = n^2(n-1)$. But $n$ and $a$ are coprime, so that $n$ divides 2, which isn't possible.

6. So since either $b$ or $n$ doesn't divide $an+b, S(n)$ isn't an integer for any $n$ by induction.

Answer 16

My proof is a slightly more advanced version of a proof that $H_n = \sum \limits_{k=1}^n \frac{1}{k} = \frac{a}{b}$ is of the form $\frac{odd}{even}$, where $n \geq 2$ and $\frac{a}{b}$ is an irreducible fraction. Namely, I prove that the denominator $b$ for $H_n$ is divisible by the highest power of $2$ not exceeding $n$ and is not divisible by any higher powers of $2$.

Let's prove the following lemma.

Lemma. Let $s \geq 1$ be an integer such that $2^s \leq n < 2^{s+1}$ and let $H_n = \frac{a}{b}$, an irreducible fraction. Then the denominator $b$ is of the form $2^{s}q$, where $q$ is odd.

Proof: by induction on $n$.

Base case: for $n = 2$ we have $H_2 = 1 + \frac{1}{2} = \frac{3}{2}$ and $b = 2^1$, where $q=1$ is odd.

Step: Suppose that it is true for $n-1$. Then we have $2$ cases:

• Case 1: $n = 2^s$. Then by induction hypothesis $H_{n-1} = \frac{c}{d}$ where $d = 2^{s-1}p$, and $p$ is odd. Then we have $$H_n = H_{n-1} + \frac{1}{n} = \frac{c}{d} + \frac{1}{n} = \frac{c}{2^{s-1}p} + \frac{1}{2^s} = \frac{2c+p}{2^{s}p}$$ Since $2c+p$ is odd (because $p$ is odd) we have (after possibly reducing the resulting fraction, but not affecting powers of $2$) that $b=2^{s}q$.

• Case 2: $2^s < n < 2^{s+1}$. Then by induction hypothesis $H_{n-1} = \frac{c}{d}$ where $d = 2^{s}p$, and $p$ is odd. Now let $n = 2^{r}t$, where $t$ is odd. Note that we have $r < s$. Then we calculate $$H_n = H_{n-1} + \frac{1}{n} = \frac{c}{d} + \frac{1}{n} = \frac{c}{2^{s}p} + \frac{1}{2^{r}t} = \frac{2^r ct + 2^{s}p}{2^{s+r}pt} = \frac{ct+2^{s-r}p}{2^{s}p}$$ Since $ct+2^{s-r}p$ is odd (because $ct$ is odd and also $s-r > 0$) we have (after possibly reducing the resulting fraction, but not affecting powers of $2$) that $b=2^{s}q$.

Now the proof of theorem is obvious. We have that for $n\geq 2$ we have $H_n = \frac{a}{b}$, an irreducible fraction with even denominator. This is impossible since $b$ must be equal to $1$ if $H_n$ is an integer.