I have asked earlier if the series $\sum_{n\geq1} \frac{2^n\bmod n} {n^2}$ is divergent. The definite answer has yet to come. I was wondering now about the easier case of $\sum_{n\geq1} \frac{P(n)\bmod Q(n)} {n^2}$ where the degree of $Q(n)$ is $1$.

  • $\begingroup$ are you taking $P,Q$ with integer coefficients and doing the mod operations on the integers ? $\endgroup$ – mercio Oct 12 '16 at 12:13
  • $\begingroup$ @ Mercio They can be real coefficients! $\endgroup$ – Aurelian Florea Oct 12 '16 at 12:14
  • $\begingroup$ so you can take $P(n) = n$ and $Q(n) = 2n$ and this gives a divergent series ? $\endgroup$ – mercio Oct 12 '16 at 12:15
  • $\begingroup$ Probably there isn't a general answer for all P and Q. I'm interested if there is an answer for specific cases. I'm not sure it is that easy! $\endgroup$ – Aurelian Florea Oct 12 '16 at 12:21
  • $\begingroup$ @mercio: My understanding was that the OP performs $P \mod Q$ as polynomials, applying the result in $n$ (so that $P(n) \mod Q(n)$ is just an abuse of notation for $(P \mod Q) (n)$. My interpretation is based on the OP's words "the remainder is a always a constant" which is compatible with his specification of $\deg Q = 1$. The OP should clarify this. $\endgroup$ – Alex M. Oct 12 '16 at 12:28

Notice that

$$ P(n) \text{ mod } Q(n) = Q(n) \bigg\{ \frac{P(n)}{Q(n)} \bigg\} $$

From this, by multiplying $-1$ to both $P$ and $Q$ if needed, we may assume that the leading coefficient of $Q$ is positive. Also, since $\deg Q = 1$, we can write

$$ P(x) = A(x)Q(x) + B $$

for some polynomial $A(x)$ and some constant $B$. Thus

$$ \bigg\{ \frac{P(n)}{Q(n)} \bigg\} = \bigg\{ A(n) + \frac{B}{Q(n)} \bigg\}. $$

Now we consider several cases:

  • Assume that the sequence $(A(n) : n \geq 1)$ takes integer values only. In particular, this forces that $A(x) \in \Bbb{Q}[x]$, and in fact, Pólya classified all such polynomials. Then for large $n$ we have $\{P(n)/Q(n)\} = B/Q(n)$. Thus in this case, $$ \sum_{n=1}^{N} \frac{P(n) \text{ mod } Q(n)}{n^2} = \sum_{n=1}^{N} \frac{B}{n^2} + O(1) \quad \text{as } N \to \infty $$ and the summation converges.

  • Assume that $(\{A(n)\} : n \geq 1)$ is periodic and $A(n) \notin \Bbb{Z}$ for some $n$. Then there exist positive integers $a, b \geq 1$ and sufficiently small $\epsilon > 0$ such that $ \{A(ak+b)\} \geq \epsilon. $ Then for large $k$ we have $\{ P(ak+b)/Q(ak+b)\} \geq \epsilon $ as well and thus $$ \sum_{n=1}^{\infty} \frac{P(n) \text{ mod } Q(n)}{n^2} \geq \sum_{k=1}^{\infty} \frac{P(ak+b) \text{ mod } Q(ak+b)}{(ak+b)^2} \geq \epsilon \sum_{k=1}^{\infty} \frac{Q(ak+b)}{(ak+b)^2} + O(1). $$ Comparing the last term with the harmonic series, the summation diverges.

  • Finally, assume that $(\{A(n)\} : n \geq 1)$ is never periodic. It can be proved that this happens exactly when the coefficient of some non-constant term of $A(x)$ is irrational. Now it is known that for such polynomial, $(\{A(n)\} : n \geq 1)$ is equidistributed. (See this.) Since $B/Q(n) \to 0$ as $n\to \infty$, it then follows that $\{ P(n)/Q(n) \}$ is also equidistributed. Thus by writing $$ \sum_{n=1}^{\infty} \frac{P(n) \text{ mod } Q(n)}{n^2} = \sum_{n=1}^{\infty} \frac{Q(n)}{n^2} \bigg\{ \frac{P(n)}{Q(n)} \bigg\}, $$ we can show that the series diverges. (Try summation by parts and compare the resulting summation with the harmonic series.)

Summarizing and utilizing Pólya's classification, we reach the following conclusion:

Conclusion. The series $\sum_{n=1}^{\infty} \frac{P(n) \text{ mod } Q(n)}{n^2}$ with $\deg Q = 1$ converges if and only if $P$ is of the form $$ P(x) = Q(x)\sum_{k=0}^{n} c_k \binom{x}{k} + B $$ for some integers $c_k$ and a real number $B$.

  • $\begingroup$ I have deleted the misleading text, thanks for pointing that out! $\endgroup$ – Aurelian Florea Oct 18 '16 at 11:32

Below is a partial answer, valid for the case when $Q$ is a monic polynomial (for simplicity, I shall take its leading coefficient to be $1$, but the proof is essentially the same for $-1$).

In order for your question to make sense, I shall assume $P$ and $Q$ with integer coefficients:

$$P = \sum _{i = 0} ^d a_i X^i, \quad Q = X - p .$$

Applying Euclidean division (this is where we use that $Q$ is monic, otherwise the division should be performed in $\Bbb Q[X]$ because $\Bbb Z[X]$ is not a Euclidean ring) to the polynomials $P$ and $Q$, there exist polynomials $F$ and $R$ in $\Bbb Z[X]$ such that

$$\tag{*} P = QF + R ,$$

with $\deg R < \deg Q = 1$ (which means that $R$ is in fact just an integer number).

Evaluating (*) in $p$ gives

$$P(p) = \underbrace {Q(p)} _{=0} F(p) + R = R .$$

Evaluating (*) in $n$ gives

$$P(n) = Q(n) F(n) + R = Q(n) F(n) + P(p) ,$$

which shows that

$$P(n) \mod Q(n) = P(p) \mod Q(n) .$$

Now, if $n$ is such that $P(p) < Q(n)$ (which means $n > p + P(p)$), then

$$P(p) \mod Q(n) = P(p) .$$

Putting all these together it follows that

$$\sum _{n = 1} ^\infty \frac {P(n) \mod Q(n)} {n^2} = \sum _{n = 1} ^\infty \frac {P(p) \mod Q(n)} {n^2} = \\ \sum _{n = 1} ^{P(p) + p - 1} \frac {P(p) \mod Q(n)} {n^2} + \sum _{n = P(p) + p} ^\infty \frac {P(p) \mod Q(n)} {n^2} = \\ \sum _{n = 1} ^{P(p) + p - 1} \frac {P(p) \mod Q(n)} {n^2} + \sum _{n = P(p) + p} ^\infty \frac {P(p)} {n^2} .$$

The first part of the last line is just a sum with a finite number of terms, therefore finite, and the second is a convergent series, essentially because $\sum \limits _{n = 1} ^\infty \dfrac 1 {n^2}$ is well known to be convergent.

  • $\begingroup$ @ Alex M. I am in doubt as here is an example where wolfram alpha doesn't know <wolframalpha.com/input/?i=sum(((n%5E3%2B3*n%5E2%2B15*n%2B27)modulo(9n%2B17))%2F(n%5E2))> $\endgroup$ – Aurelian Florea Oct 12 '16 at 12:13
  • $\begingroup$ @AurelianFlorea: The problem is that WolframAlpha does not understand your query (it fails to parse it). I recommend that you use the language of the Mathematica software package to input queries, and surround "atomic" symbols by spaces, like I do. $\endgroup$ – Alex M. Oct 12 '16 at 12:25
  • $\begingroup$ Actually This is not what I asked. I'm not sure if the polynomial remainder applies. I meant for example wolframalpha.com/input/… this. I used the polynomial division as a hint but I'm not sure it works. I meant the number division between n^3+3n^2+15*n+27 and 9n+7 whichever n was supposed to be. Like in the example with 2^n. I'm sorry if I cause any misshapenings! $\endgroup$ – Aurelian Florea Oct 12 '16 at 13:10
  • $\begingroup$ I'm sorry for causing trouble, It is just my second question, I'm still learning! $\endgroup$ – Aurelian Florea Oct 12 '16 at 14:05
  • $\begingroup$ @AurelianFlorea: I have written a new (partial) answer, that reflects the intended meaning of your question. $\endgroup$ – Alex M. Oct 12 '16 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.