Let $0<\alpha\leq\frac{\pi^2}{6}.\ $ Does $\exists\ A\subset\mathbb{N}$ and $f:A\to\{-1,1\}$ such that $\sum_{n\in A} \frac{f(n)}{n^2}=\alpha?$

Question

If we let $0<\alpha\leq \frac{\pi^2}{6},\ $ then it is not always true that $\exists\ A\subset \mathbb{N}$ such that $\displaystyle\sum_{n\in A} \frac{1}{n^2} = \alpha.\ $ To see this, consider the fact that $\ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645,\ $ and $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}-1 \approx 0.645,\ $ and so there is no $\ A\subset \mathbb{N}$ such that $\displaystyle\sum_{n\in A} \frac{1}{n^2} = 0.9,\ $ which is $<1$ but $>0.645.$

This prompted me to ask the following:

Let $0<\alpha\leq \frac{\pi^2}{6}.\ $ Does $\exists\ A\subset \mathbb{N}$ and $f:A\to \{-1,1\}$ such that $\displaystyle\sum_{n\in A} \frac{f(n)}{n^2} = \alpha?$

Answer 1

The answer is yes.

Milo Brandt's answer to a related question (which was pointed out by Martin R in a comment on the OP) contains this lemma:

Lemma: Let $s_n$ be any sequence of nonnegative real numbers with the property that $s_n \leq \sum_{i=n+1}^{\infty}s_n$ and $\lim_{n\rightarrow\infty}s_n = 0$. Then, for any $0\leq x \leq \sum_{i=1}^{\infty}s_n$, there is some subset $S\subseteq \mathbb N$ such that $x=\sum_{i\in S}s_i$.

We can then establish the OP's conjecture in parts:

• If $x\in[0,\frac{\pi^2}6-1]$, then apply the above lemma with $(s_n) = (\frac1{n^2})_{n=2}^\infty$.
• If $x\in[2-\frac{\pi^2}6,1]$, then apply the above lemma to $1-x$, again with $(s_n) = (\frac1{n^2})_{n=2}^\infty$. Then subtracting that representation from $\frac1{1^2}$ yields an appropriate representation of $x$ itself.
• If $x\in[1,\frac{\pi^2}6]$, then apply the above lemma to $x-1$, again with $(s_n) = (\frac1{n^2})_{n=2}^\infty$. Then adding that representation to $\frac1{1^2}$ yields an appropriate representation of $x$ itself.

(Note that the intervals in the first two cases overlap and thus together cover all of $[0,1]$.)

By a similar argument, it can be shown that one can always choose $f$ to take the value $-1$ at most twice (and this is best possible for $x\in(\frac{\pi^2}6-1,\frac34)$).

Answer 2

We first note:

Lemma. For sequences of reals $\mathbf{a} = (a_n)_{n\in\mathbb{N}}$ and $\mathbf{b} = (b_n)_{n\in\mathbb{N}}$, define $ L_{\mathbf{a}}(\mathbf{b})$ by

$$ L_{\mathbf{a}}(\mathbf{b}) = \sum_{n\in\mathbb{N}} b_n a_n = \lim_{\substack{F \uparrow \mathbb{N} \\ F : \text{finite}}} \sum_{n \in F} b_n a_n $$

whenever the limit exists in $[-\infty, \infty]$. If $a_n \to 0$ and $a_{n+1} \geq \frac{1}{2}a_n > 0$ for all $n\in\mathbb{N}$, then

$$ L_{\mathbf{a}}(\{0, 1\}^{\mathbb{N}}) = [0, L_{\mathbf{a}}(\mathbf{1})], $$

where $\mathbf{1} = (1)_{n\in\mathbb{N}}$ is the constant sequence with value $1$ so that $L_{\mathbf{a}}(\mathbf{1}) = \sum_{n\in\mathbb{N}} a_n$.

Its proof can be found in my previous answer.

Now let $\mathbf{a} = (\frac{1}{n^2})_{n\geq 1}$. Since the condition $a_{n+1} \geq \frac{1}{2} a_n > 0$ holds for all $n \geq 3$ for this sequence, the above lemma tells that

\begin{align*} L_{\mathbf{a}}(\{ \mathbf{b} \in \{0, 1\}^{\mathbb{N}} : b_1 = 0\}) &= L_{\mathbf{a}}(\{ \mathbf{b} : b_1 = 0, b_2 = 0\}) \cup L_{\mathbf{a}}(\{ \mathbf{b} : b_1 = 0, b_2 = 1\}) \\ &= [0, \zeta(2) - \tfrac{5}{4}] \cup [\tfrac{1}{4}, \zeta(2)-1] \\ &= [0, \alpha], \end{align*}

where $\alpha = \zeta(2) - 1 \approx 0.64493$. From this, it is easy to find that

\begin{align*} L_{\mathbf{a}}(\{ \mathbf{b} \in \{-1, 0, 1\}^{\mathbb{N}} : b_1 = 0\}) &= [-\alpha, \alpha]. \end{align*}

(First show that $L_{\mathbf{a}}(\mathbf{b})$ for $\mathbf{b}$ with $b_1 = 0$ and $b_n \in \{-1, 0, 1\}$, $n \geq 2$ is bounded between $-\alpha$ and $\alpha$. Then use the above result to show that $L_{\mathbf{a}}(\mathbf{b})$ can indeed take any values in $[-\alpha, \alpha]$.) Therefore

\begin{align*} L_{\mathbf{a}}(\{-1, 0, 1\}^{\mathbb{N}}) &= \bigcup_{c \in \{-1, 0, 1\}} L_{\mathbf{a}}(\{ \mathbf{b} \in \{-1, 0, 1\}^{\mathbb{N}} : b_1 = c\}) \\ &= \bigcup_{c \in \{-1, 0, 1\}} [c - \alpha, c + \alpha] \\ &= [-\zeta(2), \zeta(2)]. \end{align*}