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Let's change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

At least for $\alpha$ a quadratic algebraic irrational, we have affirmative answer. In this case, by diophantine approximation, there is $C>0$ such that $\lvert\alpha-p/q\rvert>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz–Szász theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!

Let's change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

[useless assumption] At least for $\alpha$ a quadratic algebraic irrational, we have affirmative answer. In this case, by diophantine approximation, there is $C>0$ such that $\lvert\alpha-p/q\rvert>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz–Szász theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!

Let's we change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

At least for $\alpha$ a quadratic algebraic irrational, we have affermative answer. In this case, by diophantine approximation, there is $C>0$ such that $|\alpha-p/q|>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz-Szasz theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!

Let's change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

At least for $\alpha$ a quadratic algebraic irrational, we have affirmative answer. In this case, by diophantine approximation, there is $C>0$ such that $\lvert\alpha-p/q\rvert>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz–Szász theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!

Let's we change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

At least for $\alpha$ a quadratic algebraic irrational, we have affermative answer. In this case, by diophantine approximation, there is $C>0$ such that $|\alpha-p/q|>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz-Szasz theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

Let's we change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

At least for $\alpha$ a quadratic algebraic irrational, we have affermative answer. In this case, by diophantine approximation, there is $C>0$ such that $|\alpha-p/q|>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz-Szasz theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!