1 Introduction
Let
$\mathbb {Z}$
,
$\mathbb {Z}^+$
and
$\mathbb {Q}$
be the set of integers, the set of positive integers and the set of rational numbers, respectively. Let
$n\in \mathbb {Z}^+$
. More than one hundred years ago, Theisinger [Reference Theisinger9] proved that the nth harmonic sum
$1+\tfrac 12+\cdots + \frac{1}{n}$
is an integer if and only if
$n=1$
. Eight years later, Nagell [Reference Nagell7] extended Theisinger’s theorem from the sequence of positive integers to general arithmetic progressions by showing that if a and b are positive integers and
$n\ge 1$
, then the reciprocal sum
$\sum _{i=0}^{n-1}{1}/{(a+bi)}$
is an integer if and only if
$n=a=1$
. Then Erdős and Niven [Reference Erdős and Niven2] generalised Nagell’s theorem by establishing a similar result on the integrality of the elementary symmetric functions of
${1}/{a}, {1}/{(a+b)}, \ldots , {1}/{(a+(n-1)b)}$
. More recently, the result of Erdős and Niven was extended to arbitrary polynomial sequences (see [Reference Chen and Tang1, Reference Hong and Wang4, Reference Luo, Hong, Qian and Wang6, Reference Wang and Hong10]). Very recently, Nagell’s theorem [Reference Nagell7] was extended by Feng et al. [Reference Feng, Hong, Jiang and Yin3] to the sum of reciprocals of arbitrary powers of the elements of arithmetic progressions.
Another interesting theorem obtained by Erdős and Niven in [Reference Erdős and Niven2] states that no two partial sums of the harmonic series can be equal. We extend this theorem from harmonic series to the series
$\sum _{k=0}^\infty {1}/{(a+bk)}$
. Let a and b be positive integers. For nonnegative integers m and n with
$m\le n$
, define the partial sum,
$$ \begin{align*}H_{a, b}(m, n) :=\sum\limits_{k=m}^{n}\frac{1}{a+kb}.\end{align*} $$
The main result of this paper can be stated as follows.
Theorem 1.1. Let a and b be positive integers. Then no two partial sums of the series
$\sum _{k=0}^\infty {1}/{(a+bk)}$
can be equal. That is,
$H_{a, b}(m_1, n_1)=H_{a, b}(m_2, n_2)$
if and only if
$(m_1, n_1)=(m_2, n_2)$
.
As usual, for any prime p and integer m, we let
$\nu _p(m)$
stand for the p-adic valuation of m, that is,
$\nu _p(m)$
is the biggest nonnegative integer r with
$p^r$
dividing m. If
$x={m_1}/{m_2}\in \mathbb {Q}$
, where
$m_1$
and
$m_2$
are integers and
$m_2\ne 0$
, then we define
$\nu _p(x):=\nu _p(m_1)-\nu _p(m_2)$
.
This paper is organised as follows. In Section 2, we supply some preliminary lemmas which are needed in the proof of Theorem 1.1. Then in Section 3, we give the proof of Theorem 1.1, using analytic and p-adic methods. Section 4 is devoted to some remarks.
2 Auxiliary lemmas
In this section, we present two auxiliary lemmas that are needed in the proof of Theorem 1.1. We begin with the following remarkable result that is due to Shorey and Tijdeman. It plays an important role in the proof of Theorem 1.1.
Lemma 2.1 [Reference Shorey, Tijdeman, Baker, Bollabás and Hajnal8].
Let
$a, b$
and t be positive integers with
$b\ge 2$
,
$\gcd (a, b)=1$
and
$t\ge 3$
. Then the greatest prime factor
$P(a, b, t)$
of the product
$\prod _{k=0}^{t-1}(a+kb)$
is no less than t. Moreover,
$P(a, b, t)=t$
if and only if
$(a, b, t)=(2, 7, 3)$
.
Remark 2.2. It is clear that
$\gcd (P(a, b, t), b)=1$
and there is exactly one term in
$\{a+kb\}_{k=0}^{t-1}$
that is divisible by
$P(a, b, t)$
as
$t\le P(a, b, t)$
and the set
$\{a+kb\}_{k=0}^{P(a, b, t)-1}$
is a complete system of residues modulo
$P(a, b, t)$
.
The second lemma is also used by Erdős and Niven.
Lemma 2.3 [Reference Erdős and Niven2].
For any real number
$x>\tfrac 12$
,
$$ \begin{align*}\frac{1}{x}=\log \frac{x+1/2}{x-1/2} -\sum_{j=1}^{\infty}\frac{2}{(2j+1)(2x)^{2j+1}}.\end{align*} $$
3 Proof of Theorem 1.1
Assume that
$$ \begin{align} H_{a, b}(m_1, n_1)=H_{a, b}(m_2, n_2) \end{align} $$
for some positive integers
$m_1, n_1, m_2$
and
$n_2$
with
$(m_1, n_1)\ne (m_2, n_2)$
. Without loss of generality, let
$m_1\le m_2$
. By cancelling the same terms on both sides of (3.1), we can further assume that
$0\le m_1\le n_1<m_2\le n_2$
. If
$a>b$
, then
$$ \begin{align*}H_{a, b}(m, n)=H_{a-b, b}(m+1, n+1).\end{align*} $$
Also (3.1) is equivalent to
$$ \begin{align*}H_{a/g, b/g}(m_1, n_1)=H_{a/g, b/g}(m_2, n_2),\end{align*} $$
where
$g:=\gcd (a, b)$
. From these two equations and Theorem 2 of [Reference Erdős and Niven2], we can assume that
$0<a<b, b\ge 2$
and
$\gcd (a, b)=1$
in what follows.
First, because
$H_{a, b}(m_1, n_1)>H_{a, b}(m_2, n_2)$
if
$n_1-m_1\ge n_2-m_2$
, by (3.1) one must have
$n_2-m_2>n_1-m_1\ge 0$
. From (3.1),
$$ \begin{align*}(n_1-m_1+1)a^{-1} \equiv \sum\limits_{k=m_1}^{n_1}\frac{1}{a+kb} =\sum\limits_{k=m_2}^{n_2}\frac{1}{a+kb} \equiv (n_2-m_2+1)a^{-1} \operatorname{mod} b,\end{align*} $$
where
$a^{-1}$
stands for an integer such that
$a\cdot a^{-1} \equiv 1 \operatorname {mod} b$
(which exists because
$\gcd (a, b)=1$
). Thus
$$ \begin{align} (n_2-m_2)-(n_1-m_1)\equiv 0 \operatorname{mod} b. \end{align} $$
Because
$n_2-m_2>n_1-m_1\ge 0$
, it then follows that
$$ \begin{align*}t:=n_2-m_2+1\ge n_1-m_1+b+1\ge 0+2+1=3.\end{align*} $$
Now let p be the greatest prime factor of the product
$$ \begin{align*}\prod_{k=m_2}^{n_2}(a+kb)=\prod_{k=0}^{n_2-m_2}(a+m_2b+kb).\end{align*} $$
Then, by Lemma 2.1,
$p\ge t$
and there is exactly one term, say
$a+k_2b$
, in
$\{a+kb\}_{k=m_2}^{n_2}$
that is divisible by p. However, if
$p=t$
, then
$(a+m_2b, b, n_2-m_2+1)=(2, 7, 3)$
. It follows that
$m_2=0$
, a contradiction because
$m_2>0$
. Hence,
$$ \begin{align} p\ge t+1. \end{align} $$
In addition, by (3.1) and
$$ \begin{align*}v_p(H_{a, b}(m_2, n_2))=v_p\bigg(\frac{1}{a+k_2b}\bigg)<0,\end{align*} $$
there is at least one term in
$\{a+kb\}_{k=m_1}^{n_1}$
that is divisible by p. Because
$n_1-m_1+1<t<p$
, there is exactly one term, say
$a+k_1b$
, in
$\{a+kb\}_{k=m_1}^{n_1}$
that is divisible by p. Clearly,
$v_p(a+k_2b)=v_p(a+k_1b)$
.
Let
$a+k_1b=p^sA$
and
$a+k_2b=p^sB$
with
$s\ge 1$
being an integer and
$\gcd (p, A)=\gcd (p, B)=1$
. Because
$\gcd (a+kb, p)=1$
for
$k\in (\{l\}_{l=m_1}^{n_1} \cup \{l\}_{l=m_2}^{n_2})\setminus \{k_1, k_2\}$
, it follows by (3.1) that
$$ \begin{align*}\frac{1}{a+k_1b}-\frac{1}{a+k_2b} =\sum\limits_{\substack{k=m_2\\ k\ne k_2}}^{n_2}\frac{1}{a+kb} -\sum\limits_{\substack{k=m_1\\ k\ne k_1}}^{n_1}\frac{1}{a+kb} \in \mathbb{Z}_p,\end{align*} $$
where
$\mathbb {Z}_p$
stands for the ring of p-adic integers (see, for example, [Reference Koblitz5]). Together with the identity
$$ \begin{align*}\frac{1}{a+k_1b}-\frac{1}{a+k_2b} =\frac{1}{p^sA}-\frac{1}{p^sB}=\frac{B-A}{p^sAB},\end{align*} $$
this shows that
$p^s \mid (B-A)$
. However,
$(k_2-k_1)b=p^s(B-A)$
and
$\gcd (p, b)=1$
. So
$p^{2s} \mid (k_2-k_1)$
.
Now, from
$m_1\le k_1\le n_1$
,
$m_2\le k_2\le n_2$
and (3.3),
$$ \begin{align*}(t+1)^2\le p^2 \le p^{2s} \le k_2-k_1 \le n_2-m_1 = m_2+t-1-m_1.\end{align*} $$
So
$$ \begin{align} m_2-m_1\ge t^2+t+2, \end{align} $$
and
$$ \begin{align*}n_2\ge n_2-m_1\ge (t+1)^2=(n_2-m_2+2)^2>(n_2-m_2)^2 +(n_2-m_2).\end{align*} $$
One can immediately deduce that
$$ \begin{align} n_2<m_2+\sqrt{m_2} \end{align} $$
and
$$ \begin{align} t=n_2-m_2+1<\sqrt{m_2}+1. \end{align} $$
However, from (3.1),
$$ \begin{align} \frac{1}{a+m_1b}\le H_{a, b}(m_1, n_1) =H_{a, b}(m_2, n_2)< \frac{t}{a+m_2b} \end{align} $$
because
$m_2<n_2$
. Together with (3.4), this gives
$$ \begin{align*}t-1>\frac{a+m_2b}{a+m_1b}-1 =\frac{(m_2-m_1)b}{a+m_1b} \ge \frac{(t^2+t+2)b}{a+m_1b}.\end{align*} $$
It then follows from
$0<a<b$
and
$t\ge 3$
that
$$ \begin{align*}1+m_1>\frac{a+m_1b}{b}> \frac{t^2+t+2}{t-1}=t+2+\frac{4}{t-1}\ge 7.\end{align*} $$
Hence,
$m_1\ge 7$
and again by (3.4),
$$ \begin{align*}m_2\ge m_1+t^2+t+2\ge 21.\end{align*} $$
From Lemma 2.3,
$$ \begin{align} bH_{a, b}(m_1, n_1) & =\sum\limits_{k=m_1}^{n_1}\frac{1}{k+a/b} \nonumber \\ & =\sum\limits_{k=m_1}^{n_1}\log\frac{k+a/b+1/2}{k+a/b-1/2} -\sum\limits_{k=m_1}^{n_1}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}} \nonumber \\ & =\log\frac{n_1+a/b+1/2}{m_1+a/b-1/2} -\sum\limits_{k=m_1}^{n_1}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}}. \end{align} $$
Likewise,
$$ \begin{align} bH_{a, b}(m_2, n_2) & =\sum\limits_{k=m_2}^{n_2}\frac{1}{k+a/b} \nonumber \\ &=\log\frac{n_2+a/b+1/2}{m_2+a/b-1/2} -\sum\limits_{k=m_2}^{n_2}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}}. \end{align} $$
Let
$$ \begin{align*}N_1:=n_1+a/b+1/2, \quad N_2:=n_2+a/b+1/2\end{align*} $$
and
$$ \begin{align*}M_1:=m_1+a/b-1/2, \quad M_2:=m_2+a/b-1/2.\end{align*} $$
$$ \begin{align} \log\frac{N_1M_2}{M_1N_2} &= \sum\limits_{k=m_1}^{n_1}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}} -\sum\limits_{k=m_2}^{n_2}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}} \nonumber \\ &= \sum_{j=1}^{\infty}\frac{1}{2j+1} \bigg( \sum\limits_{k=m_1}^{n_1} \frac{2}{(2(k+a/b))^{2j+1}}-\sum\limits_{k=m_2}^{n_2} \frac{2}{(2(k+a/b))^{2j+1}}\bigg). \end{align} $$
In what follows, we establish upper and lower bounds for
$\log {N_1M_2}/{M_1N_2}$
.
On the one hand, for any positive integer j, because
$7\le m_1 \le n_1<m_2<n_2$
,
$$ \begin{align*} &\sum\limits_{k=m_1}^{n_1} \frac{2}{(2(k+a/b))^{2j+1}} -\sum\limits_{k=m_2}^{n_2}\frac{2}{(2(k+a/b))^{2j+1}} \\ &\qquad >\frac{1}{(2(n_1+a/b))^{2j}}\sum\limits_{k=m_1}^{n_1} \frac{1}{k+a/b}-\frac{1}{(2(m_2+a/b))^{2j}} \sum\limits_{k=m_2}^{n_2}\frac{1}{k+a/b}\\ &\qquad =\bigg(\frac{1}{(2(n_1+a/b))^{2j}}-\frac{1}{(2(m_2+a/b))^{2j}}\bigg) \sum\limits_{k=m_2}^{n_2}\frac{1}{k+a/b}\\ &\qquad >0. \end{align*} $$
This together with (3.10) implies that
$$ \begin{align*}N_1M_2>M_1N_2.\end{align*} $$
Because
$4b^2N_1M_2$
and
$4b^2M_1N_2$
are both positive integers,
$$ \begin{align*} &4b^2(N_1M_2-M_1N_2) = (2bN_1)(2bM_2)-(2bM_1)(2bN_2) \\ &\quad=(2a+(2n_1+1)b)(2a+(2m_2-1)b) -(2a+(2m_1-1)b)(2a+(2n_2+1)b) \\ &\quad=4ab(n_1+m_2-m_1-n_2) +b^2((2n_1+1)(2m_2-1)-(2m_1-1)(2n_2+1)) \\ &\quad\equiv 4ab(n_1+m_2-m_1-n_2) \operatorname{mod} 2b^2\\ &\quad\equiv 0 \operatorname{mod} 2b^2, \end{align*} $$
where (3.2) is applied in the last congruence. So we have
$$ \begin{align*}N_1M_2-M_1N_2\ge \frac{2b^2}{4b^2}=\frac{1}{2}.\end{align*} $$
It is clear that
$$ \begin{align*}\log(1+x)>\frac{x}{2} \quad\mbox{for } 0<x<2.\end{align*} $$
It then follows from the fact that
$M_1, M_2, N_1$
and
$N_2$
are all positive that
$$ \begin{align} \log \frac{N_1M_2}{M_1N_2} &= \log \bigg(1+\frac{N_1M_2-M_1N_2}{M_1N_2}\bigg) \nonumber \\ &\ge \log \bigg(1+\frac{1}{2M_1N_2}\bigg) \nonumber \\ &> \frac{1}{4M_1N_2} \nonumber \\ &= \frac{1}{4(m_1+a/b-1/2)(n_2+a/b+1/2)} \nonumber \\ &= \frac{1}{4((m_1+a/b)(n_2+a/b)-n_2/2+m_1/2-1/4)} \nonumber\\ &> \frac{1}{4(m_1+a/b)(n_2+a/b)}. \end{align} $$
On the other hand, because
$m_1\ge 7$
, from (3.10) and (3.1), we derive
$$ \begin{align} \log \frac{N_1M_2}{M_1N_2} &< \sum\limits_{k=m_1}^{n_1}\sum_{j=1}^{\infty} \frac{2}{(2j+1)(2(k+a/b))^{2j+1}} \nonumber \\ &< \sum\limits_{k=m_1}^{n_1}\frac{2}{3(2(k+a/b))^3} \sum_{j=1}^{\infty}\frac{1}{(2(k+a/b))^{2j-2}} \nonumber \\ &< \sum\limits_{k=m_1}^{n_1}\frac{1} {12(k+a/b)^3}\sum_{j=1}^{\infty}\frac{1}{(2(7+0))^{2j-2}} \nonumber\\ &= \sum\limits_{k=m_1}^{n_1}\frac{1}{12(k+a/b)^3}\cdot \frac{196}{195} \nonumber \\ &< \frac{49}{585}\cdot \frac{1}{(m_1+a/b)^2} \sum\limits_{k=m_1}^{n_1}\frac{1}{k+a/b} \nonumber \\ &= \frac{49}{585}\cdot \frac{1}{(m_1+a/b)^2} \sum\limits_{k=m_2}^{n_2}\frac{1}{k+a/b} \nonumber \\ &< \frac{49}{585}\cdot \frac{1}{(m_1+a/b)^2}\cdot\frac{t}{m_2+a/b}. \end{align} $$
$$ \begin{align*}\frac{1}{4(m_1+a/b)(n_2+a/b)} <\frac{49}{585}\cdot \frac{t}{(m_1+a/b)^2 (m_2+a/b)}.\end{align*} $$
Hence, by (3.7), (3.5), (3.6) and
$m_2\ge 21>16$
,
$$ \begin{align*} 2.98\ldots =\frac{585}{196} &<\frac{t(n_2+a/b)}{(m_1+a/b)(m_2+a/b)} \\ &<\frac{t^2(n_2+a/b)}{(m_2+a/b)^2} \\ &<\frac{(\sqrt{m_2}+1)^2(1+m_2+\sqrt{m_2})} {m_2^2} \\ &<\frac{(\sqrt{m_2}+1)^4}{m_2^2} \\ &=\bigg(1+\frac{1}{\sqrt{m_2}}\bigg)^4 \\ &<\bigg(1+\frac{1}{\sqrt{16}}\bigg)^4=\frac{625}{256}=2.44\ldots , \end{align*} $$
which is impossible. This concludes the proof of Theorem 1.1.
4 Final remarks
(1) Note that
$\tfrac 12+\tfrac 13+\tfrac 14 \equiv \frac{1}{12} \operatorname {mod} 1$
and
$\tfrac 12+\tfrac 13+\tfrac 14 +\tfrac 15+\tfrac 16\equiv \tfrac 14+\tfrac 15 \operatorname {mod} 1$
. Let a and b be positive integers. Clearly
$H_{1,b}(0,n) \equiv H_{1,b}(1,n) \bmod 1$
for any positive integer
$n$
. Whether the congruence
$H_{a, b}(m_1, n_1)\equiv H_{a, b}(m_2, n_2) \operatorname {mod} 1$
holds for infinitely many values
$m_1, n_1, m_2, n_2$
with
$a+m_1b \ne 1$
and
$a+m_2b \ne 1$
is an unsolved problem.
(2) Associated to a polynomial
$f(x)$
with nonnegative integer coefficients and
$f(0)\ne 0$
, one can form an infinite sequence
$\mathcal {S}_f:= \{f(n)\}_{n=0}^{\infty }$
of positive integers. A partial sum of the series
$\sum _{k=0}^\infty {1}/{f(k)}$
is defined by
$$ \begin{align*}H_{f}(m, n) :=\sum\limits_{k=m}^{n}\frac{1}{f(k)},\end{align*} $$
where
$0\le m\le n$
. A natural question arises. For a given polynomial
$f(x)$
with nonnegative integer coefficients and
$f(0)\ne 0$
, is it true that
$$ \begin{align*}H_{f}(m_1, n_1)=H_{f}(m_2, n_2)\end{align*} $$
holds if and only if
$(m_1, n_1)=(m_2, n_2)$
?
Theorem 1.1 gives a positive answer to this question when
$\deg {f}=1$
. However, the answer to this question may be negative when
$\deg {f}\ge 2$
. Here is an example. Let
$f(x)=x^{2t}+2^t$
with t being a positive integer. Then
$$ \begin{align*}\frac{1}{f(0)}=\frac{1}{2^t}=\frac{1}{1+2^t}+ \frac{1}{2^{2t}+2^t}=\frac{1}{f(1)}+\frac{1}{f(2)}.\end{align*} $$
Acknowledgment
The authors wish to thank the anonymous referee for careful reading of the manuscript and helpful comments and suggestions.
