Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer?

Question

Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer? I have no clue how to begin. I've tried to simplify this somehow, but with no effect.

Answer 1

Let $X = (1+a)(1+2a)\cdots(1+na)$

Let $Y = \sum_{k = 1}^{n}\frac{X}{1+ka}$

Let $Z = \frac{Y}{X}$

Let $p^r$ be the highest power of a prime $p$ that divides $X$

Let $p^s$ be the highest power of $p$ that divides $1+ka$

So the power of $p$ that divides $Y$ is at least $p^r/p^s = p^{r-s}$

So $p^r$ which divides $X$ does not divide $Y \implies Z$ is not an integer.

Answer 2

Extended proof

Here I extend the proof of my original post for $a = 2$ to arbitary integers $a > 2$.

Notice that I consider the proof incompete as it makes an assumption which I could not prove (but which was used by myself and others before).

Let

$$c(k)=1 + a \;k$$ $$m=\sum _{k=0}^n \frac{1}{c(k)}$$ $$x=\prod _{k=0}^n c(k)$$

Assumption

Let $p^r$ be the highest power of a prime $p$ which divides $x$.

In the following we assume that there is a unique maximum, i.e. we rule out the case that there are two (or more) primes in $x$ which have the same "highest" power $r$

Hence the numbers $p$ and $r$ are uniquely defined for each $a$ and $n$.

Now we extract possible powers of $p$ from the $c(k)$ writing

$$c(k) = p^{s(k)} q(k)$$

where

$$p\nmid q(k)\tag{1}$$

Hence

$$m=\sum _{k=0}^n \frac{1}{p^{s(k)} q(k)}\tag{2}$$

Here we can assume without loss of generality that the order of the summands is such that

$$r = s(0) < s(1) <= ... <= s(n)\tag{3}$$

Now multiplying $(2)$ by $p^r q(0)$ gives

$$m p^r q(0) = 1 + \sum _{k=1}^n \frac{p^{r-s(k)} q(0) }{q(k)}\tag{4}$$

Taking this relation mod $p$, observing $(3)$ and $q(k) \ne 0 \;mod(p)$ due to $(1)$, leads to the contradiction

$$0 = 1$$

which completes the proof (if the assumption is correct).

Original post

Proof for a = 2.

I adapt the beautiful proof by Bill Dubuque in Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

Proof by contradiction: suppose the sum of the inverted first n odd integers is an integer m.

Since there is a unique denominator $\rm\:\color{#C00} {3^K}\:$ having maximal power of $3,\,$ upon multiplying all terms through by $\rm\:3^{K-1}$ one deduces the contradiction that $\rm\ 1/3\, =\, c/d \;$ with $\rm\ 3\nmid d$, $ $ e.g.

$$\begin{eqnarray} & &\rm\ \ \ \ \color{green}{m} &=&\ \ 1 &+& \frac{1}{3} &+& \frac{1}{5} &+& \frac{1}{7} &+&\color{#C00}{\frac{1}{9}} &+& \frac{1}{11} &+& \frac{1}{13} &+& \frac{1}{15} \\ &\Rightarrow\ &\rm\ \ \color{green}{3m} &=&\ \ 3 &+&\ 1 &+& \frac{3}{5} &+& \frac{3}{7} &+&\, \color{#C00}{\frac{1}{3}} &+& \frac{3}{11} &+& \frac{3}{13} &+& \frac{3}{15}^\phantom{M^M}\\ &\Rightarrow\ & -\color{#C00}{\frac{1}{3}}\ \ &=&\ \ 3 &+&\ 1 &+& \frac{3}{5} &-&\rm \color{green}{3m} &+& \frac{3}{11} &+& \frac{3}{13} &+& \frac{1}{5}^\phantom{M^M} \end{eqnarray}$$

The sum on the r.h.s. has only denominators which have no factor 3 so reduces to a fraction with a denominator $d$ with $3\nmid d$. This contradicts the l.h.s.

Answer 3

If there is a prime (or, for that matter a prime power) dividing only one of the denominators, then the sum is not an integer. Now by the prime number theorem in arithmetic progressions, for any fixed $a$ this holds for sufficiently large $n$, as there will be a prime $p = 1 + k a$ for some $k$ with $n/2 < k \le n$. I don't know if there is a good enough result to work for every $a$ and $n$.

EDIT: I tried all $a$ from $2$ to $100$ and $n$ from $2$ to $10000/a$. In all these cases there was at least one prime that divided only one of the denominators. Of course that is not a proof that it works for all $n$ and $a$.

Answer 4

This is not a complete answer (yet). I'm posting it because it may be useful to people who are thinking about this problem.

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Note that

$$\gcd(1+a,1+ka)=\gcd(1+a,(k-1)a)=\gcd(1+a,k-1)$$

Let $p\mid a+1$ (where $p$ is prime). For $\frac1{1+a}+\frac1{1+2a}+\cdots+\frac1{1+na}$ to be an integer (we leave the $1+$ at the start away since that's an integer already), we need

$$(1+a)(1+2a)\cdots(1+na)\ \ \mid \ \ \sum_{k=1}^n\prod_{\substack{1\le i\le n\\ i\neq k}} (1+ai)$$

Since $p$ divides the left-hand side, it must divide the right-hand side; so,

$$p\ \ \mid \ \ \sum_{k=1}^n\prod_{\substack{1\le i\le n\\ i\neq k}} (1+ai)$$

But each of the terms on the left has a $1+a$ in it, except when $k=1$; so, $p\mid \prod_{i=2}^n(1+ai)$, which means $p\mid 1+ai$ for some $2\le i\le n$. But we've seen that $\gcd(1+a,1+ia)=\gcd(1+a,k-1)$, so that $p\mid i-1$ for some $2\le i\le n$. Now we can do the same trick with $1+2a$ instead of $1+2a$; let $q\mid 1+2a$, similarly derive that $q\mid 1+ia$ for some $i\in\{1,3,4,\cdots,n\}$, which means $q\mid i-2$. This we can do for every $1+ka$; we obtain:

Let $p$ be a prime such that $p\mid 1+ak$. Then there exists a number $i\in\{1,2,\cdots,n\}\backslash\{k\}$ such that $p\mid i-k$.

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This tells us that the numbers $1+a,1+2a,\cdots,1+na$ all need to have prime factors smaller than $n$.

Let now $p<n$ be a prime such that $p\not\mid a$. Then we know for $i=1,2,\cdots,p$, that all $1+ia$ are distinct $\mod p$ (since $1+ai\equiv 1+aj\mod p$ implies $i\equiv j\mod p$). This in turn means there's a $1\le i\le p$ such that $1+ai\equiv 0\mod p$, so that $p$ is a factor of $1+ai$.

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We've now determined exactly what the prime factors of $(1+a)(1+2a)\cdots(1+na)$ are; they are all primes less than $n$ that don't divide $a$.