Given a non-negative sequence $a_n$, strictly decreasing and tending to zero, can we show that (for large $n$)
$$ \frac{a_n - a_{n+1}}{a_n} \approx \frac{a_n}{na_n} = \frac{1}{n} \text{ }?$$
Obviously, we are inspired to attempt this by noticing that $a_n - a_{n+1} \to l \implies \frac{a_n}{n} \to l$, since the latter is (roughly) the Cesaro mean of the former (i.e., the Cesaro mean of the sequence $b_n = a_n - a_{n+1}$). If we had $l \neq 0$, then we could just note that the quotient of these two sequences would head towards 1, which would quickly give the result. However, I can't think of a way to handle this for the case where the sequence $a_n$ tends to zero.
Examples ($\frac{1}{n}, \frac{1}{\sqrt[3]{n}}, \frac{\log(n)}{n},$ etc.) suggest that the claim is true.
Try something that converges faster, like $a_n = 2^{-n}$.
Or more generally, for any $f$ with $0 < f(n) < 1$ and $c > 0$, you can define a sequence satisfying your conditions by
$$ a_0 = c \qquad \qquad a_{n+1} = a_n (1 - f(n)) $$
that has
$$ \frac{a_n - a_{n+1}}{a_n} = f(n) $$