# Are all partial consecutive harmonic subsums distinct?

Let $b \gt a \geq 0$ be integers, and as elsewhere let $H_n$ be $\sum^n_{i=1} 1/i$. A partial consecutive harmonic subsum is a number $H(a,b)$ of the form $H_b - H_a$ (with $H_0=0$). If $c=a$ and $d=b$ are two other integers we have of course $$H(a,b)=H(c,d).$$ Question: Are there any other cases where we have equality , especially where $d \gt c \gt b \gt a$ ? Note that $d=c$ and $b=a$ Is ruled out.

The following shows a connection with prime numbers. If $p$ is a prime number with $2p \gt b \geq p \gt a$, then $p$ is a factor of the denominator of the (reduced) fraction that is $H(a,b)$. There are more elaborate conditions that imply that a prime belongs to the denominator. If there is a nontrivial solution, then the (products of the numbers in the) intervals $(a,b]$ and $(c,d]$ have most if not all of their prime factors in common. In particular $(c,d]$ should have all composite numbers.

I am hoping for a distinctness result to help with a question about (non-exact) packing of the harmonic series. Again, references on this question to the literature are appreciated.

Edit 2017.02.05 GRP:

I appreciate the comments and links offered so far. Inspired by a suggestion of Włodzimierz Holsztyński, I offer a conjecture toward his comparison of $H(ka,kb)$ and $H(na,nb)$. The idea is to break both sums into $nk$ many pieces and compare the partial sums. Toward showing $H(na,nb) \gt H(ka,kb)$ for integers $n \gt k \gt 0$, I conjecture the following possibly stronger condition. Given integers $b \gt 0$ and $n \gt k \gt 0$, for all $0 \lt j \leq nk$ the following inequality holds: $$\sum^j_{i=1} \frac1{k(bn + \lceil i/k \rceil)} \gt \sum^j_{i=1} \frac1{n(bk + \lceil i/n \rceil)} .$$

I would be pleased to see this resolved. End Edit 2017.02.05 GRP

Gerhard "Maybe Tie-in With Grimm Later" Paseman, 2017.01.30.

• For $c\leq 10^7$ and $d=c+e$ with $3\leq e\leq 10$ I looked for solutions to $1/(a+1)+1/(a+2)=H_d-H_c$; this is not hard because it just involves checking to see if a quadratic with integer coefficients has integer solutions. I didn't find any solutions. – Kevin Buzzard Jan 30 '17 at 20:00
• I suspect no such solution pairs exist. No proof though. – T. Amdeberhan Jan 30 '17 at 21:32
• Since $H(a,b) = H(c,d) \Longleftrightarrow H(a,c) = H(b,d)$, we can restrict to $a < b \le c < d$. My alternative argument in the answer below shows that when $b \le c$ and $H(a,b) = H(c,d)$, then there can be no prime $p$ with $c < p \le d$, which is a fairly strong condition (it forces $H(c,d)$ to be quite small, for example, but we don't know how small exactly, I believe). Unfortunately, I don't quite see how to develop this idea into a proof... – Michael Stoll Jan 31 '17 at 12:02
• "If c=a and d=b are two other integers"--pardon me? – Włodzimierz Holsztyński Feb 2 '17 at 5:04
• Did somebody check the proof given in one of the answers on MathSE? – Michael Stoll Feb 2 '17 at 13:40

Here is a partial result:

If we fix $a < b$, then there are at most finitely many $(c,d)$ such that $H(a,b) = H(c,d)$.

First note that, from the asymptotics, $d-c$ must get large with $c$. Now consider the 2-adic valuation of $H(a,b)$; say $H(a,b) = 2^{-v} r$ with $r$ a 2-adic unit. Then for $c$ large enough, the interval $[c+1,d]$ contains a unique integer with maximal 2-adic valuation $> v$, and so the valuation of $H(c,d)$ is strictly less than $-v$. "Large enough" means that $d-c \ge 2^{v+1}$ (then one of the numbers in the interval is divisible by $2^{v+1}$, so the maximal 2-adic valuation is $> v$).

(Alternatively, one can use the Prime Number Theorem. When $c$ is large, $[c+1,d]$ contains a prime. Let $p$ be the largest such prime; then $2p > d$ (Bertrand's postulate), so the denominator of $H(c,d)$ is divisible by $p$, but the denominator of $H(a,b)$ is not, as long as $c \ge b$. This is related to a remark in the question.)

• To conclude that the denominator of $H(a,b)$ is not divisible by $p$, we need to assume that $d/2>b$, not that $c\ge b$, correct? – Seva Jan 30 '17 at 16:42
• @Seva If $c \ge b$, then $p > b$, so the condition is sufficient. Of course, $d/2 > b$ implies $p > b$ as well. Neither of these will imply the other in general (which one is "better" depends on $d/c$). – Michael Stoll Jan 30 '17 at 16:46
• @MichaelStoll, thank you for your answer, and for communicating with me. – Włodzimierz Holsztyński Feb 3 '17 at 12:54
• @WłodzimierzHolsztyński You are welcome. – Michael Stoll Feb 3 '17 at 23:07

Silberger, Silberger, and Silberger, Finite segments of the harmonic series, preprint of 27 October 2011, available here, define $$\sigma(m,k)=\sum_{j=0}^{k-1}{1\over m+j}$$ and conjecture that (as a function on pairs of natural numbers) it is one-one. They prove some special cases.

There is an announcement by Donald Silberger, same title, dated 19 May 2012, available here, claiming a proof of this conjecture. The announcement promises to put the proof up on arxiv, but I didn't find it there.

I thank Gerry Myerson. His post put me on watch for papers coauthored by Silberger, of which a recent one (http://arxiv.org/abs/1702.01316) refers to a paper of Erdos and Niven (Some properties of partial sums of the harmonic series) from 1946 which answers the title question affirmatively. Thanks also to the Silbergers and Erdos and Niven.

I thank Michael Stoll for his interest in the problem and for clarifying some statements I made in the post.

I also thank Włodzimierz Holsztyński for inspiring the following proof sketch, which solves the version in the edit. (I begin the sketch now, and will finish it and provide another in a subsequent edit.) The motivation is to show for integers $0 \lt k \lt n$ and $0 \lt a \lt b$ that $H(ka,kb) \neq H(na,nb)$. Using $a=0$ and $b=1$ for guidance, I show $H(ka,kb) \lt H(na,nb)$ for $b=a+1$, which will imply the motivated result.

I start by splitting each of the $k$ summands of $H(ka,ka+k)$ into $n$ equal parts, and similarly the $n$ summands of $H(na,na+n)$ each into $k$ equal parts. That suggests comparing two sums each of $nk$ terms. Except that I am using $a$ instead of $b$, and that I want $j=nk$, the sum comparison looks exactly like the comparison of the question edit above. Of course, if I prove the conjecture in the edit, I get the result I want.

I change notation now replacing $a$ and $a+1$ with $b$ and $b+1$, and subtract the two sums term by term. It is now enough for me to show that the following holds for $0 \lt j \leq nk$:

$$\sum^j_{i=1} \frac{n \lceil i/n \rceil - k \lceil i/k \rceil}{k(nb + \lceil i/k \rceil)n(kb + \lceil i/n \rceil)} \gt 0$$

This is a straightforward induction on $j$ when $t_j = n \lceil j/n \rceil - k \lceil j/k \rceil$ is positive. Since $t_1$ is positive, the partial sum can be shown positive also when $t_j$ is zero and the induction holds. It remains to show that the sum is still positive for those $j$ for which $t_j$ is negative. This will be provided in the edit to come.

Edit 2017.02.07 GRP: To handle the case when $t_j$ is negative, I prove something even stronger. I restrict the indices $i$ and $j$ to range between $mn+1$ and $mn+n$ for an integer $m$ with $0 \leq m \lt k$, and I show that this portion (the sum for $i=mn+1$ to $j$) is also positive. I look for the largest $i$ in this range with $t_i \geq 0$ and note that it is not far from $mn+n$. If $r =(mn+n) \bmod k$ with $0 \leq r \lt k$ then this largest $i$ is $mn+n-r$. We now have $(n-r)$ positive terms of weight $r$ or more against $j-(n-r) \leq r$ negative terms of weight at most $(k-r)$ (I am suppressing mention of the denominators and the exact numerators of the summands), and since $n \gt k$, this means the positive terms outweigh the negative terms. Thus the subsubsums are also positive, so the subsums are positive, giving the desired conclusion and the end of this sketch.

I told you that story to tell you this one. The just completed sketch shows promise in giving an elementary proof of the title problem in a similar way. The hitch is that an additional term shows in the numerator. I will start this sketch and hope to finish it in a subsequent edit.

Recall that it suffices to show that subsums sharing no terms are disjoint. Now choose positive integers $a$ and $x$, and then choose $b \geq a+x$. There will be a unique integer $y \gt x$ such that $H(b,b+y-1) \lt H(a,a+x) \leq H(b,b+y)$. I want to show $H(a,a+x) \lt H(b,b+y)$ again by splitting the $1/(a+i)$ summands into $y$ equal pieces, and similarly each $1/(b+i)$ in $x$ equal pieces. I then line up the sums of the $xy$ many pieces and hope to prove positivity of the difference by proving positivity of the subsums as in the previous sketch.

When I do this, I get the following as a summand, where I use $d=ya-xb$: $$\frac{d+ y\lceil i/y \rceil -x\lceil i/x \rceil}{y(a+ \lceil i/y \rceil)x(b + \lceil i/x \rceil)} .$$ Now if $d \geq 0$, the sketch above goes through and we can celebrate. If $d$ is negative and small, I think a variant of the above sketch will work, right down to the subsubsum portion. However I do not have that to put in this edit.

If $y/b$ is small (and indeed I can use Bertrand to make it less than 1 and other results in prime gaps to make it less than 1/5) then I can show $d$ has small size (perhaps less than $(y-x)/2$), and I may be able to tweak the sketch to provide a new and weaker proof of the Erdos Niven result. I would like to make it even more elementary, so with this edit I invite others to play with it. End Edit 2017.02.07 GRP

Gerhard "Serializing Answers Seems Somewhat Easier" Paseman, 2017.02.07.

Since $\ H_n\$ is aproximately $\ \log(n) + \gamma,\$ we have $\ H(a\ b)\$ approximetely equal to $\ \log(b)-\log(a)\$ (say, when $\ b-a\$ is large when compared to $\ a>0).\$ However, for arbitrary integers $\ 0<a<b\$ we have:

$$H(a\,\ b)\ \ne\ H(2\!\cdot\!a\ \ \ 2\!\cdot\!b)$$

This is very simple. What about other inequalities, are they much harder:

$$H(k\!\cdot\!a\ \ \ k\!\cdot\!b)\ \ne\ H(n\!\cdot\!a\ \ \ n\!\cdot\!b) \qquad\qquad ?$$

A comparison between the left and the right side of the inequality may possibly help to solve the whole conjecture (could you do it before Donald Silberger's arXive article shows up? :-) )

EDIT  A stronger hypothesis than Gerhard's conjecture (or question):

CONJECTURE   If $\ a\ b\ c\ d\,\ \ k\ l\ m\ n\$ are non-negative integers such that $\ (a\ b) \ne (c\ d), \ a<b,\ c<d,\ \gcd(k\ l) = \gcd(m\ n)=1,\ H(a\ b) =\frac kl\$ and $\ H(c\ d) = \frac mn\$ then $$l\ne n.$$

• More precisely, $H_n = \log(n) + \gamma + 1/(2n) + O(n^{-2})$, so $H(a,b) \approx \log(b/a) - 1/2(1/a-1/b)$, which basically already implies that $d/c$ must be a bit smaller than $b/a$ if $H(a,b) = H(c,d)$ and $c > a$. So I don't think that proving $H(a,b) \neq H(ka,kb)$ for $k \ge 2$ is going to help a lot. – Michael Stoll Feb 2 '17 at 11:20
• @MichaelStoll, thank you for bringing here more precise estimates of $\ H_n.\$ Overall and ultimately you may be right that my proposition for helping to solve the conjecture may be not adequate. I only want to clarify (just in case) that I meant comparisons--i.e. estimations of the differences--between the two sides of the above inequalities, and not simply the fact that we (possibly) had inequalities. – Włodzimierz Holsztyński Feb 2 '17 at 12:22
• For the latest,many counter examples exist, one with c=19 and d=20. There may even be counterexamples with d-c larger than 1, but I don't know of them. Yet. (How about d=21?) Gerhard "The Future Can Be Tricky" Paseman, 2017.02.09. – Gerhard Paseman Feb 9 '17 at 15:27
• Gerhard, would you write the d=20 example explicitly? (Perhaps, you could insert it into my "Answer", noting there that it is your subtext). One could consider the whole class of such examples as the first step toward your conjecture. – Włodzimierz Holsztyński Feb 9 '17 at 21:09
• Let a=3, b=5, then H(a,b)=9/20. Similarly, H(19,20)=1/20. Of more interest is 3,7 and 19,21, with l=n=420. Gerhard "See How Easy It Is?" Paseman, 2017.02.09. – Gerhard Paseman Feb 9 '17 at 21:30