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clarify remarks about other solutions etc.
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Gareth McCaughan
Gareth McCaughan
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The "analytic process"

Start with the empty set of integers, yielding LHS=0 and RHS=1. Now repeatedly do the following: pick the smallest positive integer that (1) hasn't been used already and (2) preserves the condition LHS < RHS.

On the face of it this isn't a very explicit construction -- obviously you can just do it but it's not clear what the result looks like. But in fact it turns out we can understand it fairly well.

The "algebraic process"

WeGeneralizing a bit

The things I've (rather dubiously) called "analytic" and "algebraic" are just two ways of looking at the exact same process, of course. But if we do something slightly different to one of them, it may have no equivalent in terms of the other.

Specifically, we can get infinitely many sequences by picking larger numbers than thesemodify the "analytic" process so as to give (as promised at the start) "plenty of solutions" for $N=\infty$. Suppose we perform that process as described except that some finite number of times we pick a larger number than it asks for. Everything will still work, and then continuing with the process as described abovewe'll get an "$N=\infty$" solution, but it will no longer be describable in nice "algebraic" terms. This gives onlyus countably infinitely many "$n=\infty$$N=\infty$" solutions. If we want more, but unless I'm confused Ithen (I think -- I haven't thought it through super-carefully) we can also just always pickdo the "analytic" process except that at each step we use either the smallest permitted $n$number it asks for or the number 1 more than thisbigger, and everythingagain we will still work, givingalways get the LHS and RHS converging to a common value. So there are continuum-many solutions.

Alternatively, we could modify the "algebraic" process: start with some different $p_1,q_1$ and perform the calculations it calls for. This will give us different initial values of $L,R$, and the final result will have the property that $L+\sum 1/n=R\cdot\prod(1-1/n)$. This will not (except by some monstrous coincidence) be a solution to the original problem, but it might be of interest anyway; for instance, if we choose $p_1=3,q_1=2$ then we have $L=0,R=\frac12$, so this gives us a solution to $\sum 1/n=\frac12\prod(1-1/n)$.

Start with the empty set of integers, yielding LHS=0 and RHS=1. Now repeatedly do the following: pick the smallest positive integer that (1) hasn't been used already and (2) preserves the condition LHS < RHS.

On the face of it this isn't a very explicit construction -- obviously you can just do it but it's not clear what the result looks like. But in fact it turns out we can understand it fairly well.

We can get infinitely many sequences by picking larger numbers than these some finite number of times and then continuing with the process as described above. This gives only countably many "$n=\infty$" solutions, but unless I'm confused I think we can also just always pick either the smallest permitted $n$ or 1 more than this, and everything will still work, giving continuum-many solutions.

The "analytic process"

Start with the empty set of integers, yielding LHS=0 and RHS=1. Now repeatedly do the following: pick the smallest positive integer that (1) hasn't been used already and (2) preserves the condition LHS < RHS.

On the face of it this isn't a very explicit construction -- obviously you can just do it but it's not clear what the result looks like. But in fact it turns out we can understand it fairly well.

The "algebraic process"

Generalizing a bit

The things I've (rather dubiously) called "analytic" and "algebraic" are just two ways of looking at the exact same process, of course. But if we do something slightly different to one of them, it may have no equivalent in terms of the other.

Specifically, we can modify the "analytic" process so as to give (as promised at the start) "plenty of solutions" for $N=\infty$. Suppose we perform that process as described except that some finite number of times we pick a larger number than it asks for. Everything will still work, and we'll get an "$N=\infty$" solution, but it will no longer be describable in nice "algebraic" terms. This gives us countably infinitely many "$N=\infty$" solutions. If we want more, then (I think -- I haven't thought it through super-carefully) we can do the "analytic" process except that at each step we use either the number it asks for or the number 1 bigger, and again we will always get the LHS and RHS converging to a common value. So there are continuum-many solutions.

Alternatively, we could modify the "algebraic" process: start with some different $p_1,q_1$ and perform the calculations it calls for. This will give us different initial values of $L,R$, and the final result will have the property that $L+\sum 1/n=R\cdot\prod(1-1/n)$. This will not (except by some monstrous coincidence) be a solution to the original problem, but it might be of interest anyway; for instance, if we choose $p_1=3,q_1=2$ then we have $L=0,R=\frac12$, so this gives us a solution to $\sum 1/n=\frac12\prod(1-1/n)$.

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Gareth McCaughan
Gareth McCaughan
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I can at any rate answer Q3: Yes, there are (plenty of) solutions for "N=infinity". And it will turn out that answering this in more detail resolves at least part of Q2: there is at least one solution for each positive N>1.

Start with the empty set of integers, yielding LHS=0 and RHS=1. Now repeatedly do the following: pick the smallest positive integer that (1) hasn't been used already and (2) preserves the condition LHS < RHS.

There always is such an integer, since large enough x will make the changes in LHS,RHS as small as you like. So the process goes on for ever.

The LHS is increasing and the RHS is decreasing. Any LHS value is a lower bound for all subsequent RHSes, and any RHS value is an upper bound for all subsequent LHSes. So the LHS and RHS are bounded monotone sequences and hence have limits.

I claim the limits are equal; if $R-L$ is bounded below by $\delta$ then any $n>1/\delta$ is acceptable for condition 2, so eventually we will use all such $n$ -- but these $n$ on their own are enough to make the LHS diverge to infinity and the RHS to zero, contradiction.

On the face of it this isn't a very explicit construction -- obviously you can just do it but it's not clear what the result looks like. But in fact it turns out we can understand it fairly well.

Suppose at some stage we have partial sum $L$ and partial product $R$. The next $n$ will be the smallest integer bigger than $(R+1)/(R-L)$. Write $p=(R+1)/(R-L)$ and $q=1/(R-L)$. I claim that in fact $p,q$ are always integers!

(Note that this will also give us solutions for each $N$: take $n=p$ and the next $L,R$ will be equal.)

So, initially we have $L=0$ and $R=1$ so $p,q=2,1$: integers as required.

Now suppose $p,q$ are integers as claimed. We'll work out what happens next and see that the "next" values of $p,q$ are integers too. Specifically, let's begin by getting $L,R$ in terms of $p,q$: we have $L=(p-q-1)/q$ and $R=(p-q)/q$.

If $p,q$ are integers then our next $n$ is $p+1$, so we replace $L$ with $L'=L+1/(p+1)$ and $R$ with $R'=p/(p+1)\cdot R$. In terms of $p,q$ this turns out to mean $L'=(p^2-pq-1)/(p+1)q$ and $R'=(p^2-pq)/(p+1)q$.

This means $R'-L'=1/(p+1)q$ and so $p'=(R'+1)/(R'-L')=p^2+q$ and $q'=1/(R'-L')=(p+1)q$. And if $p,q$ are integers then so are $p',q'$.

So, to summarize:

  • Write $p_1,q_1=2,1$ and $p_{n+1},q_{n+1}=p_n^2+q_n,(p_n+1)q_n$.
  • We get a solution for "$n=\infty$" by taking our integers to be the $p_n+1$. We get a solution for any finite $n$ by taking an initial subsequence of the $p+1$ and then, finally, the next $p$.
  • The resulting partial sum and product are $(p_n-q_n-1)/q_n$ and $(p_n-q_n)/q_n$.
  • The differences RHS-LHS are $1/q_n$.
  • The common limit of LHS and RHS (in the "$n=\infty$" case) is the limit of $p_n/q_n-1$.

We can get infinitely many sequences by picking larger numbers than these some finite number of times and then continuing with the process as described above. This gives only countably many "$n=\infty$" solutions, but unless I'm confused I think we can also just always pick either the smallest permitted $n$ or 1 more than this, and everything will still work, giving continuum-many solutions.