Is there a $f: \mathbb{N} \to \mathbb{N}$ such that $\sum_{n=1}^{\infty} \frac{1}{n^2f(n)} \in \mathbb{Q}$?

Question

Take by convention $0 \not \in \mathbb{N}$, and let $f: \mathbb{N} \to \mathbb{N}$. Define the real number $N(f)$ by

$$N(f) = \sum_{n=1}^{\infty} \frac{1}{n^2f(n)}.$$

$N(f)$ is well-defined because, for example, each summand is a positive number less than $\frac{1}{n^2}$, so $N(f) \leq \frac{\pi^2}{6}$, so we have a bounded above sequence of positive numbers.

My question: is $N(f)$ ever rational?

Note that if it is the case that $N(f)$ is never rational, we're almost certainly not going to be able to prove it in this thread. By taking $f(n)=n$ we obtain $\zeta(3)$, which was only proven to be irrational in the 1970s, and apparently "at least one of $\zeta(5), \zeta(7), \zeta(11),\zeta(13)$ must be irrational" (see link), so I assume the irrationality of these numbers individually is unknown.

But I'm wondering if I'm missing some $f$ for which it is not too hard show is a counterexample to the conjecture "$N(f)$ is never rational."

What I've tried thinking about is defining $f$ recursively, but not gotten far. If we say $a_k, b_k$ are defined such that $$\frac{a_k}{b_k} = \sum_{n=1}^{k} \frac{1}{n^2f(n)}$$ is in lowest terms, then we know

$$\frac{a_{k+1}}{b_{k+1}} = \frac{a_k}{b_k} + \frac{1}{(k+1)^2f(k+1)} = \frac{a_k (k+1)^2f(k+1) + b_k}{b_k (k+1)^2f(k+1)},$$

so in particular $da_{k+1} = a_k (k+1)^2f(k+1) + b_k$, $db_{k+1} = b_k (k+1)^2f(k+1)$, where $d$ is the gcd of the numerator and denominator. If we can pick $f(k+1)$ in such a way to make $d$ large, we can have the denominators of the partial sums grow slowly, which may at least be a starting point for looking for such an $f$, I'm not sure.

Answer 1

Yes, there is an $f$ such that $N(f)$ is rational. Using $$ \sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6, \\ \sum_{n=1}^\infty\frac1{(2n+1)^2(2n-1)}=\frac34-\frac{\pi^2}{16} $$ (the second one is computed using partial fractions), we have a chance to cancel the $\pi^2$ stuff.

Say, take $f(n)=2$ for $n$ even, $f(n)=3n-6$ for $n>1$ odd, and $f(1)$ arbitrary.

Answer 2

Here is a solution with $f$ a polynomial function: if $f(n)=(n+1)(n+2)^2$, then $N(f)=\frac1{16}$.

This may be proven by telescoping: \begin{align*} \sum_{n=1}^\infty\frac 1{n^2(n+1)(n+2)^2} &=\sum_{n=1}^\infty\left(\frac{1}{4n^2(n+1)^2}-\frac1{4(n+1)^2(n+2)^2}\right)\\ &=\frac1{4\cdot 1^2\cdot 2^2}=\frac1{16}. \end{align*}

Answer 3

For every positive real number $x \le \pi^2/6$ exists a function $f: \Bbb N\to \Bbb N$ with $N(f) = x$, in particular for all rational numbers in that range.

Such a function $f$ can be defined recursively with a “greedy algorithm”: For every $n$, $f(n)$ is chosen as small as possible with $$ \sum_{k=1}^n \frac{1}{k^2 f(k)} < x \, . $$

This is a special case of the following more general result:

Let $\sum_{n=1}^\infty a_n$ be a convergent series of positive real numbers with the property that $$ \tag{$*$} \sum_{k=n+1}^\infty a_k \ge \frac 12 a_n $$ for all $n \in \Bbb N$, and $x \in \Bbb R$ with $0 < x \le \sum_{n=1}^\infty a_n$. Define $(f(n))_{n \ge 1}$ and $(x_n)_{n \ge 0}$ recursively as follows: $x_0 = x$, and $$ f(n) = 1 + \left\lfloor \frac{a_n}{x_{n-1}} \right\rfloor \, , \, x_n = x_{n-1} - \frac{a_n}{f(n)} \, . $$ for $n \ge 1$. Then $\sum_{n=1}^\infty \frac{a_n}{f(n)} = x$.

$a_n = 1/n^2$ satisfies the condition $(*)$:
$$ \sum_{k=n+1}^\infty \frac{1}{k^2} \ge \sum_{k=n+1}^\infty \frac{1}{k(k+1)} = \sum_{k=n+1}^\infty \left( \frac 1k - \frac{1}{k+1}\right) = \frac{1}{n+1} \ge \frac{1}{2n^2} $$ since $2n^2 \ge n+1$.

Examples: For $a_n = 1/n^2$ and $x=1$ the algorithm produces $$ f(n) = 2, 1, 1, 1, 1, 1, 3, 9, 171, 122014, 17661589931, \ldots \, , $$ which is A377205 in the On-Line Encyclopedia of Integer Sequences. For $x=1/16$ one gets $$ f(n) = 17, 69, 2086, 3670318, 8621587552516, \ldots \, , $$ which is different from $f(n) = (n+1)(n+2)^2$ in Carl's answer. This demonstrates that different functions $f$ can have the same value $N(f)$.

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Proof: First note that $$ f(n) > \frac{a_n}{x_{n-1}} \implies \frac{a_n}{f(n)} < x_{n-1} $$ so that both $(x_n)_{n \ge 0}$ and $(f(n))_{n \ge 1}$ are well-defined. Also $(x_n)$ is strictly decreasing and strictly positive. We will now show that $$ \tag{$**$} x_n \le \sum_{k=n+1}^\infty a_k $$ for all $n \ge 0$. This implies $x_n \to 0$, and therefore $$ \sum_{n=1}^N \frac{a_n}{f(n)} = \sum_{n=1}^N (x_{n-1} - x_n) = x - x_N \to x $$ for $N \to \infty$.

Proof of $(**)$ with induction: The base case is $x_0 = x \le \sum_{n=1}^\infty a_n$, which is true by assumption. Now assume that $n \ge 1$ and $$ x_{n-1} \le \sum_{k=n}^\infty a_k \, . $$ We distinguish three cases:

• Case 1: $x_{n-1} > a_n$. Then $f(n) = 1$ and $$ x_n = x_{n-1} - a_n \le \sum_{k=n+1}^\infty a_k \, . $$

• Case 2: $a_n/2 < x_{n-1} \le a_n$. Then $f(n) = 2$ and $$ x_n = x_{n-1} - \frac{a_n}{2} \le \frac{a_n}{2} \overset{(*)}{\le} \sum_{k=n+1}^\infty a_k \, . $$

• Case 3: $x_{n-1} \le a_n/2$. Then $$ x_n < x_{n-1} \le \frac{a_n}{2} \overset{(*)}{\le} \sum_{k=n+1}^\infty a_k \, . $$

The completes the proof of $(**)$ and therefore the proof of the main result.

Answer 4

Carl's result can be slightly generalized, in the sense that the sum

$$\sum_{n=1}^{\infty} \frac{1}{n^2(n+k)(n+2k)^2}= \frac{1}{4k}\sum_{n=1}^{\infty} \left(\frac{1}{n^2(n+k)^2}-\frac{1}{(n+k)^2(n+2k)^2}\right)$$ is rational if $k\in \{1,2,\ldots\}$.

$\bf{Added:}$ @Martin R pointed out that the actual value of the sum is $$\frac{1}{4k} \sum_{n=1}^k\frac{1}{n^2(n+k)^2}$$

(so clearly rational), a particular case of

$$\sum_{n=1}^{\infty} (g(n)-g(n+k)) = \sum_{n=1}^k g(n)$$

if $g(n) \to 0$ as $n\to \infty$ ( and $k\ge 1$ an integer).

$\bf{Added:}$ Found a method to produce more rational sums where the denominator is a polynomial with integral roots. For instance, this will get

$$\sum_{n=1}^{\infty} \frac{1}{n^2(n+2)(n+5)^2(n+15)} = \frac{36145891}{52702650000}$$

the idea being that a rational fraction $$R(x) = \frac{Q(x)}{(x+a_1)^{e_1} \cdots (x+a_k)^{e_k}}$$

with $a_i \in \mathbb{N}$, and $Q(x)$ rational coefficients.

will produce a rational sum $\sum_{n=1}^{\infty} R(n)$ if in the partial fractions decomposition

$$R(x) = \sum_s \left( \sum_{\ell=1}^k \frac{c_{\ell s}}{(x+a_{\ell})^s}\right)$$

we have $\sum_{\ell=1}^k c_{\ell s}= 0$ for all $s\ge 1$. So now we are looking for $c_{\ell, s}$ that produce a fraction with numerator of degree $0$. Now this imposes a linear system on the $c_{\ell,s}$ with number of equations $=$ number of unknowns and we need the determinant of the system to be $0$. So this imposes a condition on the $a_{\ell}$'s and now it is a search for integral ( rational) solutions. Once such an $(a_{\ell} )$ ( with a prescribed $e_l$) is found, one simply takes

$$R(n) = \frac{1}{(n+a_1)^{e_1} \cdots (n+a_k)^{e_k}}$$

and this will have a rational sum.

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the answer here seems relevant, wonder if there are rational fractions $R(x)\in \mathbb{Q}(x)$ with $\sum R(n) $ a rational number and $R(x) \ne S(x+1) - S(x)$ for another rational fraction.

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$\bf{Added:}$ Found an improvement in the method to look for rational sums. It starts directly from the partial fractions expansion and then produces equation. For example, consider the partial fraction expansion of a fraction with simple roots at the denominators

$$\frac{1}{(x+a)(x+b)(x+c)(x+d)} = -\frac{1}{(a - b) (a - c) (a - d) (a + x)} -\frac{ 1}{(b-a)(b-c)(b-d) (b + x)} -\frac{1}{(c-a)(c-b)(c-d) (c + x)} -\frac{1}{(d-a)(d-b)(d-c) (d + x)}$$

Now, to find the partial fractions expansion of $\frac{1}{(x+a)^2(x+b)^2(x+c)(x+d)}$ take partial derivatives of the above equality with respect to $a$ and to $b$. Now, the terms containing the $\frac{1}{(x+a)^2}$ and $\frac{1}{(x+b)^2}$ must have the sum of coefficients $=0$. We get the equality :

$$a^2 + b^2 - a c - b c - a d - b d + 2 c d=0$$

Notice that this equality is invariant under $(a,b,c,d) \mapsto (a+r, b+r, c+r, d+r)$, as it should. Find rational solutions of the above equation ( simple, since it is solvable wr to $c$ ( or $d$), then make them integral, and then natural. For instance $(0,5,2,15)$ is a solution.

Other similar fractions produces systems that seem hard to handle $$\frac{1}{(x^2(x+b)^2(x+c)^2(x+d)(x+e)}$$ (we took $a=0$ ) we get the condition $$b^4 c^2 - 2 b^3 c^3 + b^2 c^4 - b^4 c d + b^3 c^2 d + b^2 c^3 d - b c^4 d + b^3 c d^2 - 2 b^2 c^2 d^2 + b c^3 d^2 - b^4 c e + b^3 c^2 e + b^2 c^3 e - b c^4 e + 2 b^4 d e - 2 b^2 c^2 d e + 2 c^4 d e - 2 b^3 d^2 e + b^2 c d^2 e + b c^2 d^2 e - 2 c^3 d^2 e + b^3 c e^2 - 2 b^2 c^2 e^2 + b c^3 e^2 - 2 b^3 d e^2 + b^2 c d e^2 + b c^2 d e^2 - 2 c^3 d e^2 + 2 b^2 d^2 e^2 - 2 b c d^2 e^2 + 2 c^2 d^2 e^2=0$$

Can one tell whether such an equation has rational solutions $(b,c,d,e)$ with distinct non-zero components? There was another case that I reduced to an elliptic curve having just some given rational points. But this gets even more involved.

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With a bit more effort I managed to find infinitely many rational sums of the form

$$\sum_{n=1}^{\infty}\frac{1}{(n+a)^3(n+b)^3(n+c)(n+d)(n+e)(n+f)}$$

one example being

$$(a,b,c,d,e,f) = (0, 7285919617600, 732239475268561, 1077798760000, 3603081081600, 8794728633600)$$

The method is as above, expand the fraction $\frac{1}{(n+a)^3(n+b)^3(n+c)(n+d)(n+e)(n+f)}$ and postulate that the sum of the coefficients of $1/(n+a,b,\ldots)^k$ are $0$ for $k=2,3$ ( for $k=1$ they will always be). This system will have a parametrization for solutions, not that hard), then look for something where $a$ is the smallest component.

Note: Mathematica would not evaluate the last series, I guess the numbers are just too large. Any ideas for this?

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@Carl Schildkraut found another beautiful example!! (see his comment below the answer)

$$\sum_{n\ge 2} \frac{1}{n^3(n+2)^3(n-1)^2(n+3)^2}=\frac{5}{62208}$$

following up on that: the fraction

$$R(x) = \frac{1}{(x+a)^3(x+b)^3(x+c)^2(x+d)^2}$$

(with $a$, $b$, $c$, $d$ distinct integers) will have a rational sum if the $a$, $b$, $c$, $d$ satisfy:

$$(a + b - c - d)(a^2 + b^2 - a c - b c - a d - b d + 2 c d) =0 \ \textrm{and} \\ (-2 a^7 b^3 + 8 a^6 b^4 - 12 a^5 b^5 + 8 a^4 b^6 - 2 a^3 b^7 + 3 a^7 b^2 c - 9 a^6 b^3 c + 6 a^5 b^4 c + 6 a^4 b^5 c - 9 a^3 b^6 c + 3 a^2 b^7 c - 7 a^7 b c^2 + 10 a^6 b^2 c^2 + 15 a^5 b^3 c^2 - 30 a^4 b^4 c^2 + 15 a^3 b^5 c^2 + 10 a^2 b^6 c^2 - 7 a b^7 c^2 + 3 a^7 c^3 + 12 a^6 b c^3 - 42 a^5 b^2 c^3 + 15 a^4 b^3 c^3 + 15 a^3 b^4 c^3 - 42 a^2 b^5 c^3 + 12 a b^6 c^3 + 3 b^7 c^3 - 9 a^6 c^4 + 6 a^5 b c^4 + 36 a^4 b^2 c^4 - 30 a^3 b^3 c^4 + 36 a^2 b^4 c^4 + 6 a b^5 c^4 - 9 b^6 c^4 + 9 a^5 c^5 - 20 a^4 b c^5 - a^3 b^2 c^5 - a^2 b^3 c^5 - 20 a b^4 c^5 + 9 b^5 c^5 - 3 a^4 c^6 + 9 a^3 b c^6 - 6 a^2 b^2 c^6 + 9 a b^3 c^6 - 3 b^4 c^6 + 3 a^7 b^2 d - 9 a^6 b^3 d + 6 a^5 b^4 d + 6 a^4 b^5 d - 9 a^3 b^6 d + 3 a^2 b^7 d + 8 a^7 b c d - 14 a^6 b^2 c d - 14 a^2 b^6 c d + 8 a b^7 c d - 2 a^7 c^2 d - 7 a^6 b c^2 d + 21 a^5 b^2 c^2 d + 21 a^2 b^5 c^2 d - 7 a b^6 c^2 d - 2 b^7 c^2 d + 3 a^6 c^3 d - 12 a^5 b c^3 d + 21 a^4 b^2 c^3 d + 21 a^2 b^4 c^3 d - 12 a b^5 c^3 d + 3 b^6 c^3 d + 3 a^5 c^4 d - 2 a^4 b c^4 d - 49 a^3 b^2 c^4 d - 49 a^2 b^3 c^4 d - 2 a b^4 c^4 d + 3 b^5 c^4 d - 7 a^4 c^5 d + 28 a^3 b c^5 d + 42 a^2 b^2 c^5 d + 28 a b^3 c^5 d - 7 b^4 c^5 d + 3 a^3 c^6 d - 15 a^2 b c^6 d - 15 a b^2 c^6 d + 3 b^3 c^6 d - 7 a^7 b d^2 + 10 a^6 b^2 d^2 + 15 a^5 b^3 d^2 - 30 a^4 b^4 d^2 + 15 a^3 b^5 d^2 + 10 a^2 b^6 d^2 - 7 a b^7 d^2 - 2 a^7 c d^2 - 7 a^6 b c d^2 + 21 a^5 b^2 c d^2 + 21 a^2 b^5 c d^2 - 7 a b^6 c d^2 - 2 b^7 c d^2 + 6 a^6 c^2 d^2 + 18 a^5 b c^2 d^2 - 84 a^4 b^2 c^2 d^2 - 84 a^2 b^4 c^2 d^2 + 18 a b^5 c^2 d^2 + 6 b^6 c^2 d^2 - 9 a^5 c^3 d^2 + 13 a^4 b c^3 d^2 + 56 a^3 b^2 c^3 d^2 + 56 a^2 b^3 c^3 d^2 + 13 a b^4 c^3 d^2 - 9 b^5 c^3 d^2 + 11 a^4 c^4 d^2 - 17 a^3 b c^4 d^2 + 42 a^2 b^2 c^4 d^2 - 17 a b^3 c^4 d^2 + 11 b^4 c^4 d^2 - 9 a^3 c^5 d^2 - 39 a^2 b c^5 d^2 - 39 a b^2 c^5 d^2 - 9 b^3 c^5 d^2 + 3 a^2 c^6 d^2 + 30 a b c^6 d^2 + 3 b^2 c^6 d^2 + 3 a^7 d^3 + 12 a^6 b d^3 - 42 a^5 b^2 d^3 + 15 a^4 b^3 d^3 + 15 a^3 b^4 d^3 - 42 a^2 b^5 d^3 + 12 a b^6 d^3 + 3 b^7 d^3 + 3 a^6 c d^3 - 12 a^5 b c d^3 + 21 a^4 b^2 c d^3 + 21 a^2 b^4 c d^3 - 12 a b^5 c d^3 + 3 b^6 c d^3 - 9 a^5 c^2 d^3 + 13 a^4 b c^2 d^3 + 56 a^3 b^2 c^2 d^3 + 56 a^2 b^3 c^2 d^3 + 13 a b^4 c^2 d^3 - 9 b^5 c^2 d^3 - 4 a^4 c^3 d^3 - 32 a^3 b c^3 d^3 - 168 a^2 b^2 c^3 d^3 - 32 a b^3 c^3 d^3 - 4 b^4 c^3 d^3 + 6 a^3 c^4 d^3 + 54 a^2 b c^4 d^3 + 54 a b^2 c^4 d^3 + 6 b^3 c^4 d^3 + 16 a^2 c^5 d^3 - 8 a b c^5 d^3 + 16 b^2 c^5 d^3 - 12 a c^6 d^3 - 12 b c^6 d^3 - 9 a^6 d^4 + 6 a^5 b d^4 + 36 a^4 b^2 d^4 - 30 a^3 b^3 d^4 + 36 a^2 b^4 d^4 + 6 a b^5 d^4 - 9 b^6 d^4 + 3 a^5 c d^4 - 2 a^4 b c d^4 - 49 a^3 b^2 c d^4 - 49 a^2 b^3 c d^4 - 2 a b^4 c d^4 + 3 b^5 c d^4 + 11 a^4 c^2 d^4 - 17 a^3 b c^2 d^4 + 42 a^2 b^2 c^2 d^4 - 17 a b^3 c^2 d^4 + 11 b^4 c^2 d^4 + 6 a^3 c^3 d^4 + 54 a^2 b c^3 d^4 + 54 a b^2 c^3 d^4 + 6 b^3 c^3 d^4 - 38 a^2 c^4 d^4 - 44 a b c^4 d^4 - 38 b^2 c^4 d^4 + 12 a c^5 d^4 + 12 b c^5 d^4 + 6 c^6 d^4 + 9 a^5 d^5 - 20 a^4 b d^5 - a^3 b^2 d^5 - a^2 b^3 d^5 - 20 a b^4 d^5 + 9 b^5 d^5 - 7 a^4 c d^5 + 28 a^3 b c d^5 + 42 a^2 b^2 c d^5 + 28 a b^3 c d^5 - 7 b^4 c d^5 - 9 a^3 c^2 d^5 - 39 a^2 b c^2 d^5 - 39 a b^2 c^2 d^5 - 9 b^3 c^2 d^5 + 16 a^2 c^3 d^5 - 8 a b c^3 d^5 + 16 b^2 c^3 d^5 + 12 a c^4 d^5 + 12 b c^4 d^5 - 12 c^5 d^5 - 3 a^4 d^6 + 9 a^3 b d^6 - 6 a^2 b^2 d^6 + 9 a b^3 d^6 - 3 b^4 d^6 + 3 a^3 c d^6 - 15 a^2 b c d^6 - 15 a b^2 c d^6 + 3 b^3 c d^6 + 3 a^2 c^2 d^6 + 30 a b c^2 d^6 + 3 b^2 c^2 d^6 - 12 a c^3 d^6 - 12 b c^3 d^6 + 6 c^4 d^6=0$$

Check that $(a,b,c,d) = (0,2,-1,3)$ satisfies it !

Now in the first equation the first or the second factor is $0$. Either way, we can get from that the value of $d$, and substitute in the second equation, and we get a huge simplification either way. ( happy to help with details).

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I would like to add some workaround that helps with a CAS: the formula for the coefficient $c_{k}$ for a rational fraction $R(x)$ with a polar part

$$R(x) = \frac{c_1}{(x-\alpha)^1} + \cdots + \frac{c_d}{(x-\alpha)^d} + \textrm{ other terms away from } \alpha $$ is

$$c_k = \frac{1}{(d-k)!}\left( (x-\alpha)^d R(x) \right)^{(d-k)}_{x=\alpha}$$ for all $1\le k \le d$. Before was using some ad-hoc methods and was getting bogged down in calculations (using Mathematica). This is standard, see for instance Henrici, Complex analysis.

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$\bf{Added:}$

a bit more general $$(a,b,c,d) = u(0,(2p-q)q, (2p-q)p,(p-q)q)+v$$

where $u$,$v$, $p$, $q$ are integers then

$$\sum_{n\ge n_0} \frac{1}{(n+a)^2(n+b)^2(n+c)(n+d)} \in \mathbb{Q}$$ nicest I could find being $\sum_{n\ge 1} \frac{1}{n^2(n+1)(n+3)^2(n+6)}=\frac{487}{97200}$

Answer 5

This problem is about Irrationality sequence . As noted in Wikipedia:

Hančl, Jaroslav (1991), "Expression of real numbers with the help of infinite series", Acta Arithmetica, 59 (2): 97–104

By convention $0\notin\mathbb{N}$. Given positive-integer sequence $a_n$: $$\sum_{n=1}^{\infty}\frac{1}{a_n f(n)}\in\mathbb{Q}^c\ ,\forall f:\mathbb{N\to\mathbb{N}}\implies L:=\limsup_{n\to\infty}\frac{\log\log a_n}{n}\ge\log2$$ Then, by contrapositivity: $$L:=\limsup_{n\to\infty}\frac{\log\log a_n}{n}<\log2\implies\exists f:\mathbb{N}\to\mathbb{N}\ ,\sum_{n=1}^{\infty}\frac{1}{a_nf(n)}\in\mathbb{Q}$$ In this case $a_n=n^2 \implies L=0<\log2$ , hence the existence of $f(n)$ satisfying $\sum_{n=1}^{\infty}\frac{1}{n^2f(n)}\in\mathbb{Q}$
However, finding such $f(n)$ is not an easy task (as shown in other answers).