[proofplan] We use the Euler product description of the singular series in terms of the local densities. The positivity direction is an infinite-product argument: positive factors whose deviations from $1$ are summable have a positive limiting product. The vanishing direction uses the reduction map from solutions modulo $p^j$ to solutions modulo $p^h$; if there is no solution at level $p^h$, then there is no solution at any higher level, so the corresponding local density is zero and hence the Euler product vanishes. [/proofplan] [step:Record the Euler product and the local density normalization] Let $\mathcal P$ denote the set of primes. By the assumed Euler product formula for the singular series under the stated absolute convergence hypotheses, we have \begin{align*} \mathfrak S_{s,k}(N)=\prod_{p\in\mathcal P}\sigma_p(N). \end{align*} For each $p\in\mathcal P$, the factor is normalized by \begin{align*} \sigma_p(N)=\lim_{h\to\infty}p^{h(1-s)}M_{s,k}(N;p^h). \end{align*} Here $M_{s,k}(N;p^h)$ is the number of residue classes $(x_1,\dots,x_s)\in(\mathbb Z/p^h\mathbb Z)^s$ satisfying \begin{align*} x_1^k+\cdots+x_s^k\equiv N \pmod {p^h}. \end{align*} [/step] [step:Prove that summable positive Euler factors give a positive product] [claim:Positive infinite product criterion] Let $I$ be a [countable set](/page/Countable%20Set), and let $(a_i)_{i\in I}$ be [real numbers](/page/Real%20Numbers) such that $a_i>0$ for every $i\in I$ and \begin{align*} \sum_{i\in I}|a_i-1|<\infty. \end{align*} Then the infinite product $\prod_{i\in I}a_i$ converges to a positive real number. [/claim] [proof] Choose an enumeration $I=\{i_1,i_2,\dots\}$ and define $\delta_n:=a_{i_n}-1$ for $n\in\mathbb N$. Since $\sum_{n=1}^{\infty}|\delta_n|<\infty$, there exists $n_0\in\mathbb N$ such that $|\delta_n|\le 1/2$ for every $n\ge n_0$. For $|t|\le 1/2$, the [mean value theorem](/theorems/186) applied to the map $\ell:(-1,\infty)\to\mathbb R$, $u\mapsto \log(1+u)$, on the interval with endpoints $0$ and $t$ gives \begin{align*} |\log(1+t)|\le 2|t|. \end{align*} Therefore \begin{align*} \sum_{n=n_0}^{\infty}|\log a_{i_n}|=\sum_{n=n_0}^{\infty}|\log(1+\delta_n)|\le 2\sum_{n=n_0}^{\infty}|\delta_n|<\infty. \end{align*} The finite initial sum $\sum_{n=1}^{n_0-1}\log a_{i_n}$ is finite because each $a_{i_n}$ is positive. Hence \begin{align*} L:=\sum_{n=1}^{\infty}\log a_{i_n} \end{align*} converges in $\mathbb R$. The partial products satisfy \begin{align*} \prod_{n=1}^{m}a_{i_n}=\exp\left(\sum_{n=1}^{m}\log a_{i_n}\right), \end{align*} so continuity of the exponential map gives \begin{align*} \lim_{m\to\infty}\prod_{n=1}^{m}a_{i_n}=e^L>0. \end{align*} This proves the claim. [/proof] Apply the claim with $I=\mathcal P$ and $a_p=\sigma_p(N)$ for $p\in\mathcal P$. The hypotheses $\sigma_p(N)>0$ for every $p$ and \begin{align*} \sum_{p\in\mathcal P}|\sigma_p(N)-1|<\infty \end{align*} are exactly the hypotheses of the claim. Hence \begin{align*} \prod_{p\in\mathcal P}\sigma_p(N)>0. \end{align*} Using the [Euler product identity](/theorems/1694) from the previous step, this gives \begin{align*} \mathfrak S_{s,k}(N)>0. \end{align*} [guided] We need a precise reason why infinitely many positive factors cannot multiply down to zero under the summability hypothesis. Let $I$ be a countable set and let $(a_i)_{i\in I}$ be positive real numbers with \begin{align*} \sum_{i\in I}|a_i-1|<\infty. \end{align*} Choose an enumeration $I=\{i_1,i_2,\dots\}$ and set $\delta_n:=a_{i_n}-1$. The summability condition becomes \begin{align*} \sum_{n=1}^{\infty}|\delta_n|<\infty. \end{align*} In particular, there is an integer $n_0\in\mathbb N$ such that $|\delta_n|\le 1/2$ for every $n\ge n_0$. The logarithm is the right tool because products become sums. For $|t|\le 1/2$, apply the mean value theorem to the [differentiable map](/page/Differentiable%20Map) $\ell:(-1,\infty)\to\mathbb R$ defined by $\ell(u)=\log(1+u)$. On the interval between $0$ and $t$, the derivative satisfies \begin{align*} |\ell'(u)|=\frac{1}{|1+u|}\le 2. \end{align*} Therefore \begin{align*} |\log(1+t)|\le 2|t|. \end{align*} Putting $t=\delta_n$ gives \begin{align*} \sum_{n=n_0}^{\infty}|\log a_{i_n}|=\sum_{n=n_0}^{\infty}|\log(1+\delta_n)|\le 2\sum_{n=n_0}^{\infty}|\delta_n|<\infty. \end{align*} The finitely many terms before $n_0$ cause no difficulty because each $a_{i_n}$ is positive, so each $\log a_{i_n}$ is a finite real number. Thus the real series \begin{align*} L:=\sum_{n=1}^{\infty}\log a_{i_n} \end{align*} converges. Now exponentiate the partial sums. For every $m\in\mathbb N$, \begin{align*} \prod_{n=1}^{m}a_{i_n}=\exp\left(\sum_{n=1}^{m}\log a_{i_n}\right). \end{align*} Since the partial sums of the logarithms converge to $L$ and the exponential map is continuous, the partial products converge to $e^L$. This limit is strictly positive. Applying this with $I=\mathcal P$ and $a_p=\sigma_p(N)$ proves \begin{align*} \prod_{p\in\mathcal P}\sigma_p(N)>0. \end{align*} Finally the singular series is this product by the Euler product formula, so \begin{align*} \mathfrak S_{s,k}(N)>0. \end{align*} [/guided] [/step] [step:Show that a missing solution modulo one prime power forces a zero local density] Assume that there exist a prime $p_0\in\mathcal P$ and an integer $h_0\ge 1$ such that \begin{align*} M_{s,k}(N;p_0^{h_0})=0. \end{align*} Let $j\ge h_0$ be an integer. The reduction map \begin{align*} \rho_{j,h_0}:(\mathbb Z/p_0^j\mathbb Z)^s\to(\mathbb Z/p_0^{h_0}\mathbb Z)^s \end{align*} is defined componentwise by reducing each residue class modulo $p_0^{h_0}$. If $(x_1,\dots,x_s)\in(\mathbb Z/p_0^j\mathbb Z)^s$ satisfies \begin{align*} x_1^k+\cdots+x_s^k\equiv N \pmod {p_0^j}, \end{align*} then $\rho_{j,h_0}(x_1,\dots,x_s)$ satisfies \begin{align*} x_1^k+\cdots+x_s^k\equiv N \pmod {p_0^{h_0}}. \end{align*} This would give a solution counted by $M_{s,k}(N;p_0^{h_0})$, contradicting $M_{s,k}(N;p_0^{h_0})=0$. Hence \begin{align*} M_{s,k}(N;p_0^j)=0 \end{align*} for every integer $j\ge h_0$. Using the defining normalization of the local density, we obtain \begin{align*} \sigma_{p_0}(N)=\lim_{j\to\infty}p_0^{j(1-s)}M_{s,k}(N;p_0^j)=0. \end{align*} [/step] [step:Conclude that the singular series vanishes when one local factor vanishes] From the previous step, $\sigma_{p_0}(N)=0$ for some prime $p_0$. By hypothesis, every factor $\sigma_p(N)$ is nonnegative. Therefore every finite partial product over a finite set of primes containing $p_0$ is equal to $0$. Since the Euler product defining $\mathfrak S_{s,k}(N)$ converges and is equal to the singular series, the limiting product is also $0$: \begin{align*} \mathfrak S_{s,k}(N)=\prod_{p\in\mathcal P}\sigma_p(N)=0. \end{align*} This proves both asserted criteria. [/step]