Comparing two distributions based of the ratio of their moments I am looking for some metric for distribution with support on the interval $[0,1-\epsilon]$, that will be based on the ratio of their moments.
That is, if $X\sim f(x)$, $Y\sim g(y)$, I'm looking for a metric $d(f,g)$ such that
$\frac{\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert}{\mathbb{E}X^k}$ is small for all $0<k$ $\iff$ $d(f,g)$ is small.
Of course, I could just define the distance to be the sum of these ratios, but I am not sure what this means. Can two distributions be very different but have a very close moment-ratio?
I was able to show that the 1-Wasserstein distance is small if $\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert$ is small, and vice-versa, but this is not strong enough, I want the ratio of the moments.
Is there some natural metric to look at?
 A: $\newcommand\ep\epsilon$There is no such metric, because for any $\ep\in(0,1)$ and any real $k>0$ there are random variables $X$ and $Y$ with different pdf's $f$ and $g$ supported on $[0,1-\ep]$ such that $EX^k=EY^k$.
However, the pseudometric $d_k$ defined by the formula
$$d_k(f,g):=|\ln EX^k-\ln EY^k|=\Big|\ln\frac{EY^k}{EX^k}\Big|$$
will have the desired property:
$$d_k(f,g)\to0\iff\frac{EY^k}{EX^k}\to1\iff\frac{|EY^k-EX^k|}{EX^k}\to0. \tag{1}$$

The OP has now said that all moments, for all $k>0$, were meant to be close.
Letting then
$$d(f,g):=\sum_{k=1}^\infty 2^{-k}\frac{d_k(f,g)}{1+d_k(f,g)}, \tag{2}$$
we get a metric $d$ such that
$$d(f,g)\to0\iff\frac{|EY^k-EX^k|}{EX^k}\to0\text{ for each natural }k. \tag{3}$$
Indeed, if $d(f,g)\to0$ then, by (2), for each natural $k$ we have $\dfrac{d_k(f,g)}{1+d_k(f,g)}\to0$ and hence $d_k(f,g)\to0$. Vice versa, if for each natural $k$ we have $d_k(f,g)\to0$, then $\dfrac{d_k(f,g)}{1+d_k(f,g)}\to0$ and hence, by (2) and dominated convergence,  $d(f,g)\to0$. So, $d(f,g)\to0$ if and only if for each natural $k$ we have $d_k(f,g)\to0$. Thus, (3) follows from (1).
