# Sum of the series $\sum_{n\geq1} \frac{P(n) \bmod Q(n)} {n^2}$

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 $1$.

• are you taking $P,Q$ with integer coefficients and doing the mod operations on the integers ? – mercio Oct 12 '16 at 12:13
• @ Mercio They can be real coefficients! – Aurelian Florea Oct 12 '16 at 12:14
• so you can take $P(n) = n$ and $Q(n) = 2n$ and this gives a divergent series ? – mercio Oct 12 '16 at 12:15
• 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! – Aurelian Florea Oct 12 '16 at 12:21
• @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. – Alex M. Oct 12 '16 at 12:28

## 2 Answers

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 $B$.

• I have deleted the misleading text, thanks for pointing that out! – 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.

• @ 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))> – Aurelian Florea Oct 12 '16 at 12:13
• @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. – Alex M. Oct 12 '16 at 12:25
• 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! – Aurelian Florea Oct 12 '16 at 13:10
• I'm sorry for causing trouble, It is just my second question, I'm still learning! – Aurelian Florea Oct 12 '16 at 14:05
• @AurelianFlorea: I have written a new (partial) answer, that reflects the intended meaning of your question. – Alex M. Oct 12 '16 at 15:00