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Under what conditions does the series $$\sum_ {n = 1}^\infty \frac {(f(n) +P(n)) \pmod{ Q(n)}} {D(n)}$$ converge?

Here $\text{P}$, $\text{Q}$ and $\text{D}$ are polynomials, with $\text{deg}(Q)= \text{deg}(D)-1$.

Also $f(n)= \displaystyle \sum_{r=2}^{k} a_{r} r^n$ where $ \{a_{r} \}_{r=2}^k$ are real numbers and k is a natural number with k>2. k and the degree of P must be larger than the degree of Q.

Here and here particular cases of this question have been answered by the user @SangchulLee. I'd like to know in which cases this series converges. (I assume they are rare but Sangchul Lee has already found some). I'm getting somewhere with this so don't flag me yet!

Aurelian Florea
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  • If $f(n)$ is non-constant, heuristically you should get divergence. The idea is that $f(n) + P(n)$ for large $n$ will be an exponential plus a polynomial, so taking that mod the polynomial $Q(n)$ will be chaotic enough to be approximately equidistributed in $[0, Q(n)-1]$. Take $D(n) = n^d$ and $Q(n) = n^{d-1}$ for simplicity, since that's all that's really important. The heuristic says that $(f(n) + P(n) \pmod{Q(n)})/n^{d-1}$ is equidistributed on $[0, 1]$, so half the time it'll be $\geq 1/2$. Hence $(f(n) + P(n) \pmod{Q(n)})/n^d \geq 1/(2n)$ half the time, so it'll diverge. – Joshua P. Swanson Mar 05 '21 at 09:45
  • I don't see any wiggle room for that heuristic to fail, even though proving a sufficiently precise version of it might be a mess. Since you're interested in convergence, you should focus on $f(n) = 0$. One of Sangchul's answers settles the $D(n) = n^2$ version of that question, though much of the reasoning generalizes quite readily. Have you tried running through it and seeing what breaks? – Joshua P. Swanson Mar 05 '21 at 09:47
  • No Joshua I have not, I'll try it! – Aurelian Florea Mar 05 '21 at 09:49
  • On Wolfram Alpha I have tried the case where f(n)=0 and P(n) integer coefficients and deg(Q(x))>1. that leads to A(n) to be rational and thus the series should converge. Graphically it looks that way although Wolfram Alpha cannot figure out whether the series is convergent and as a consequence it cannot output the sum. – Aurelian Florea Mar 05 '21 at 10:34
  • I doubt that f(n)>0 and P(n)>0 leads to convergent series. But maybe someone can prove it. – Aurelian Florea Mar 05 '21 at 10:35

1 Answers1

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As noted in the comments, heuristically the series will diverge if $f(n) \neq 0$. A literature search through equidistribution theorems would be the first thing to do if you wanted to prove it.

So, suppose $f(n) = 0$. Sangchul's classification for the $D(n) = n^2$ case generalizes to the present case without difficulty. The tweaks:

  • $B$ is now $B(x)$, with $\deg B(x) < \deg Q(x) < \deg D(x)$.
  • First case: $\{B(n)/Q(n)\}$ still becomes $B(n)/Q(n)$ for $n$ large and $\sum_{i=1}^\infty B(n)/D(n)$ still converges since $\deg D(x) - \deg B(x) \geq 2$.
  • Second case: the lower bound becomes $\epsilon \sum_{k=1}^\infty \frac{Q(ak+b)}{D(ak+b)} + O(1)$. Since $\deg D(x) - \deg Q(x) = 1$, this still diverges by comparison with the harmonic series.
  • Third case: the sum becomes $\sum_{n=1}^\infty \frac{Q(n)}{D(n)} \left\{\frac{P(n)}{Q(n)}\right\}$ where the term in braces remains equidistributed mod $1$. This again diverges by comparison with the harmonic series.
  • Conclusion: the series $\sum_{n=1}^\infty \frac{P(n) \pmod{Q(n)}}{D(n)}$ with $\deg Q(x) = \deg D(x) - 1$ converges if and only if $P$ is of the form $$P(x) = Q(x) \sum_{k=0}^n c_k \binom{x}{k} + B(x)$$ where $B(x)$ is a real polynomial with $\deg B(x) < \deg Q(x)$.

Hence there's no more interesting cases than the ones already found, so long as you believe the exponential heuristic.

Joshua P. Swanson
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  • I have searched for equidistribution theorems and I have found only one belonging to Weyl: https://www.math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf. Were you referring to that or something else? – Aurelian Florea Mar 05 '21 at 14:44
  • @AurelianFlorea Sure, something alone those lines anyway. Sangchul used a generalization of Weyl's result already. It seems to me you should perhaps make a new question asking about equidistribution, namely are there any cases where $(f(n) + P(n))/Q(n)$ is not equidistributed mod 1? Strictly speaking you just need a positive fraction of the time spent bounded away from 0, i.e. the only issue is if somehow the fractional part of the quotient is very biased towards very small values. Dropping the $P(n)$ for simplicity may be helpful. Note that your $f(n)$ form is important... – Joshua P. Swanson Mar 05 '21 at 22:55
  • ...as the Fibonacci example in the notes you linked demonstrates. – Joshua P. Swanson Mar 05 '21 at 22:55
  • So I'd have to prove that there is no interval included in 1..Q(n) where(f(n)+P)(n)) mod Q(n) never arrives. – Aurelian Florea Mar 06 '21 at 06:52
  • I don't think I can do it! – Aurelian Florea Mar 07 '21 at 13:53
  • Can you write the details for the general case Joshua? – Aurelian Florea Mar 08 '21 at 04:40
  • Unfortunately no, I do not know how to do the general case. Literal equidistribution is stronger than necessary, e.g. Sangchul's argument in the $2^n \pmod{n}$ case restricted to certain primes, which is a large enough fraction to still give divergence. A specific case you might want to ask if anyone can prove is whether $\sum_{n \geq 1} \left{2^n/p(n)\right}$ diverges for all polynomials $p(n)$ of degree $\geq 1$. It seems like an accessible "next step", anyway. – Joshua P. Swanson Mar 09 '21 at 07:44