We can use the Müntz–Szász theorem, which states that if $\Lambda = \{0 = \lambda_0 < \lambda_1 < \cdots\}$ then the span of $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$ if and only if $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Note: Wikipedia restricts $\Lambda \subset \mathbb N$, but this is not needed. For example see [Rudin, Real and Complex Analysis, Theorem 15.26] or [Almira, https://arxiv.org/pdf/0710.3570.pdf, Theorem 1].

Here we take $\Lambda$ to be the union of $\{0\}$ and an infinite subset of $\{\{m\alpha\} : m \in \mathbb N\}$ accumulating at $1$. The theorem then tells us that $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$.

Let $f \in C_0(\mathbb N)$. Define $F \in C[0,1]$ by $F(0) = 0$, $F(1/n) = f(n)$ for $n \in \mathbb N$, and linearly interpolating. Then we can uniformly approximate $F(x)$ by some linear combination $\sum_{i=0}^K a_i x^{\lambda_i}$. By evaluating at $0$ we find that $a_0$ must be small, so we can throw it away. Now by restricting to the values $F(1/n)$, it follows that $f(n)$ is uniformly approximated by $\sum_{i=1}^K a_i / n^{\lambda_i}$.

We can use the Müntz–Szász theorem, which states that if $\Lambda = \{0 = \lambda_0 < \lambda_1 < \cdots\}$ then the span of $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$ if and only if $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Note: Wikipedia restricts $\Lambda \subset \mathbb N$, but this is not needed. For example see [Rudin, Real and Complex Analysis, Theorem 15.26] or [Almira, Muntz type Theorems I, Theorem 1].

Here we take $\Lambda$ to be the union of $\{0\}$ and an infinite subset of $\{\{m\alpha\} : m \in \mathbb N\}$ accumulating at $1$. The theorem then tells us that $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$.

Let $f \in C_0(\mathbb N)$. Define $F \in C[0,1]$ by $F(0) = 0$, $F(1/n) = f(n)$ for $n \in \mathbb N$, and linearly interpolating. Then we can uniformly approximate $F(x)$ by some linear combination $\sum_{i=0}^K a_i x^{\lambda_i}$. By evaluating at $0$ we find that $a_0$ must be small, so we can throw it away. Now by restricting to the values $F(1/n)$, it follows that $f(n)$ is uniformly approximated by $\sum_{i=1}^K a_i / n^{\lambda_i}$.

We can use the Müntz–Szász theorem, which states that if $\Lambda = \{0 = \lambda_0 < \lambda_1 < \dots\}$ then the span of $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$ if and only if $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Note: Wikipedia restricts $\Lambda \subset \mathbb N$, but this is not needed. For example see [Rudin, Real and Complex Analysis, Theorem 15.26] or [Almira, https://arxiv.org/pdf/0710.3570.pdf, Theorem 1].

Here we take $\Lambda$ to be the union of $\{0\}$ and an infinite subset of $\{\{m\alpha\} : m \in \mathbb N\}$ accumulating at $1$. The theorem then tells us that $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$.

Let $f \in C_0(\mathbb N)$. Define $F \in C[0,1]$ by $F(0) = 0$, $F(1/n) = f(n)$ for $n \in \mathbb N$, and linearly interpolating. Then we can uniformly approximate $F(x)$ by some linear combination $\sum_{i=0}^K a_i x^{\lambda_i}$. By evaluating at $0$ we find that $a_0$ must be small, so we can throw it away. Now by restricting to the values $F(1/n)$, it follows that $f(n)$ is uniformly approximated by $\sum_{i=1}^K a_i / n^{\lambda_i}$.

We can use the Müntz–Szász theorem, which states that if $\Lambda = \{0 = \lambda_0 < \lambda_1 < \cdots\}$ then the span of $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$ if and only if $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Note: Wikipedia restricts $\Lambda \subset \mathbb N$, but this is not needed. For example see [Rudin, Real and Complex Analysis, Theorem 15.26] or [Almira, https://arxiv.org/pdf/0710.3570.pdf, Theorem 1].

Here we take $\Lambda$ to be the union of $\{0\}$ and an infinite subset of $\{\{m\alpha\} : m \in \mathbb N\}$ accumulating at $1$. The theorem then tells us that $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$.

Let $f \in C_0(\mathbb N)$. Define $F \in C[0,1]$ by $F(0) = 0$, $F(1/n) = f(n)$ for $n \in \mathbb N$, and linearly interpolating. Then we can uniformly approximate $F(x)$ by some linear combination $\sum_{i=0}^K a_i x^{\lambda_i}$. By evaluating at $0$ we find that $a_0$ must be small, so we can throw it away. Now by restricting to the values $F(1/n)$, it follows that $f(n)$ is uniformly approximated by $\sum_{i=1}^K a_i / n^{\lambda_i}$.