Looking for an example of a series $\sum_{n\ge 1} R(n)$ with $R(x) \in \mathbb{Q}(x)$ with sum in $\mathbb{Q}$, such that $R(x)\ne S(x)-S(x+1)$

Question

The title pretty much explains it, but let's provide more context:

Say we have a rational function $R(x)$ of degree $\le -2$. Then the series $\sum_{n\ge n_0} R(n)$ is convergent. The sum is very easy to evaluate if $R(x)$ happens to be a discrete difference: $$R(x) = S(x)-S(x+1)$$ for another rational fraction $S(x)$ of degree $\le -1$. There are some necessary and sufficient conditions for that, see for instance this question.

Recently there was another question on the site where some answers provided examples of series $\sum_{n\ge 1} R(n)$ with a rational sum, and $R(x)$ a rational fraction in $\mathbb{Q}(x)$. In all of the examples the sum was calculated using the fact that $R(x)$ is a discrete difference. Now the question whether there could not exist an example of a series as above with a rational sum where $R(x)$ is not a discrete difference.

Note that if the denominator of $R(n)$ has all roots integers such an example would (probably) not be possible, as this would imply some rational dependence among values of the zeta function at positive integers. We know that this is not possible for the values at even integers ( equivalent to $\pi$ is transcendental), and I do not expect that the all zeta values and $1$ would be linearly dependent over $\mathbb{Q}$. So an example would involve some other fractions. Would be happy to see an example, maybe a very simple one, that escapes me at the moment. Thank you for your interest!

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$\bf{Added:}$ Calvin Lin just provided a very ingenious answer, using zeta sums at odd integers! While the question is thus answered, one could think what else is out there... " can you surprise me with a series $\sum_{n\ge 1} R(n) \in \mathbb{Q}$ in other ways ?"

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$\bf{Added:}$ Building on the idea of Calvin Lin:

Say $T(n) = \frac{1}{n^k}$, $k>1$, then for $d\ge 2$ integer we have

$$\frac{1}{d^k} \sum_{n\ge 1} T(n) = \sum_{n\ge 1} \frac{1}{(d n)^k}=\sum_{n\equiv 0 \mod d} T(n)$$

and so $$(1-\frac{1}{d^k}) \sum_{n\ge 1} T(n) = \sum_{n\not \equiv 0 \mod d} T(n)$$

or

$$\sum_{n\ge 1}\left[ (1-\frac{1}{d^k}) T(n) - ( T(d(n-1) + 1) + T(d(n-1) + 2) + \cdot T(d(n-1) +(d-1)) ) \right] = 0$$

We could do other similar things. Let $p_1$, $p_2$, $\ldots$ $p_i$ be the prime factors of some $d\ge 2$ natural. Then we have

$$(1-\frac{1}{p_1^k})(1-\frac{1}{p_2^k}) \cdots (1-\frac{1}{p_i^k}) \sum_{n\ge 1}\frac{1}{n^k} - \sum_{(n,d) = 1} \frac{1}{n^k} = 0$$

and again we get an example. Note that we do get a series that is not a difference.

Answer 1

Verify that $$ \sum_{k=1}^\infty \frac{1}{k^2} = \frac{4}{3} + \frac{4}{3} \sum_{k=1}^\infty \frac{1}{(1+2k)^2}.$$
(Split the LHS into odd and even $k$. We can rearrange terms as the sum is finite).

Hence we have

$$ \sum_{k=1}^\infty \frac{ 1}{k^2} - \frac{ 4}{3(1+2k)^2 } = \sum_{k=1}^\infty \frac { 8k^2 + 12k + 3 } { 3k^2 ( 2k+1) ^2 } = \frac{4}{3}.$$

Then this cannot be partial fractioned into a telescoping series using terms that are a finite distance away.

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Note: Technically, I haven't definitively proved that there isn't a simple $S(x)$ of finite degree such that $\frac { 8k^2 + 12k + 3 } { 3k^2 ( 2k+1) ^2 } = S(k) - S(k+1). $
We could of course have $S(k) = \sum_{j=k}^\infty \frac { 8j^2 + 12j + 3 } { 3j^2 ( 2j+1) ^2 } $, which doesn't evaluate to a "nice form" per Eric's comment.

Answer 2

I will prove CalvinLin's sum can't be turned (directly) into a telelescoping sum. $$\frac{4}{3}=\sum_{k=1}^{\infty}\left(\frac{ 1}{k^2} - \frac{ 4}{3(2k+1)^2 }\right)=\sum_{k=1}^{\infty}R(k)$$ Where $R(k)=\frac{8k^2+12k+3}{3k^2(2k+1)^2}$. We will prove by contradiction. First, let's assume: $\ \exists S(x)\in \mathbb{Q}(x):S(x)-S(x+1)=R(x)$ (*)

Lemma 1: $\exists c\in\mathbb{R}:\lim_{x\to\infty}S(x)=c$
Proof: Since $S(x)$ is rational function, the limit is either finite or infinte (exists) and If $\lim_{x\to\infty}S(x)=\infty\implies S(x)\sim P(x)$ for some non-constant polynomial $P(x)$, hence $R(x)=S(x)-S(x+1)\sim P(x)-P(x+1)\not\to 0$, contradicts convergence of $\sum_{k=1}^{\infty}R(k)$.
The case $\lim_{x\to\infty}S(x)=-\infty$ is dealt simililarly.

Fact 2: $\forall S(x)$ (satisfies lemma 1), we can rechoose $S_1(x)=S(x)-c-4/3$ so that $\lim_{x\to\infty}S_1(x)=-4/3$

By Fact 2, we are supposed to find $S(x)$ satisfying $S(x)-S(x+1)=R(x)$ and $\lim_{x\to\infty}S(x)=S(\infty)=-4/3$, so that: $$\frac{4}{3}=\sum_{k=1}^{\infty}R(k)=\sum_{k=1}^{\infty}\left(S(k)-S(k+1)\right)=S(1)-S(\infty)=S(1)+\frac{4}{3}$$ $$\implies S(1)=0 $$ We rewrite $S(x)$ as: $S(x)=(x-1)^m G(x)$ for some positive integer $m$, and rational $G(x)$ where $G(1)$ is a finite constant.
We have $R(x)=(x-1)^m G(x)-x^m G(x+1)\ \ (2)$. From here, consider limit: $$\lim_{x\to 0}\left(R(x)-(x-1)^m G(x)\right)=\lim_{x\to 0}-x^m G(x+1)=0$$ But $R(x)\sim \frac{1}{x^2}$ as $x\to 0$, we deduce $G(x)= H_0(x)/x^2$ for some rational $H_0(x)$ defined at $0$ and $1$. Substitute back to $(2)$: $$R(x)=\frac{(x-1)^m H_0(x)}{x^2}-\frac{x^m H_0(x+1)}{(x+1)^2}$$ Now consider limit: $$\lim_{x\to -1}\left(R(x)+\frac{x^m H_0(x+1)}{(x+1)^2}\right)=\lim_{x\to -1}\frac{(x-1)^m H_0(x)}{x^2}$$ The function of limit on LHS $\sim \frac{1}{(x+1)^2}$ as $x\to -1$ hence $H_0(x)\sim \frac{1}{(x+1)^2}\implies H_{0}(x)=H_{1}(x)/(x+1)^2$ for some rational $H_{1}(x)$ defined at $1,0,-1$. We continue this argument inductively by consider limit $x\to -N$, we eventually find out that $G(x)=\frac{H_{N}(x)}{x^2(x+1)^2...(x+N)^2}$ for some rational $H_{N}(x)$ defined at $1,0,-1,...,-N$. Since $N$ is unbounded, this means $G(x)$ will have unbounded number of poles at negative integers, contradicts $G(x)\in \mathbb{Q}(x)$.

So $\not\exists S(x)$ and our initial assumption (*) is false.